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HYDROMECHANICS 

ALGER 



HYDROMECHANICS 



AN ELEMENTARY TREATISE 

Prepared for the use of the Cadets at the 
United States Naval Academy 



PHILIP R. ALGER 

Professor of Mathematics, U. S. Navy 



1902 



THE t.lB«AftY ©F 


OONGrtESS, 


Two Oopiee 


flECEfVe* 


MAR. 15 


'1902 


Cc^f^Rt»HJ 


ENTHY 


CLASS «/ XXc N«. 


COPY B. 



Copyright, 1902, by 
PHILIP R. ALGER 



THE FRIEDENWALD COMPANY 
BALTIMORE, MD., U. S. A. 






CONTENTS 



CHAPTEB PAGE 

I. Fluids Under Surface Forces 7 

II. Fluids Under Gravity 21 

III. Total Pressures — Centers of Pressure of Plane Areas. 33 

IV. Resultant Pressure on Curved Surfaces — Density and 

Specific Gravity 48 

V. Floating Bodies Continued — Stability 62 

VI. Liquids in Uniform and in Steady Motion 75 

VII. Discharge througli Small and Large Orifices 91 

VIII. Time of Emptying Vessels— Efflux of Gases— Flow 

Through Pipes 107 

IX. Compound Pipes — Water Power — Energy and Re- 
action of Jets 124 

X. Reaction and Impulse — Water Motors 141 

XL Hydraulic Machines — Work of Expanding Gases 156 

Answers to Problems 173 



CHAPTER I. 

Fluids Under Surface Forces. 

(1) Hydromechanics is the science which treats of 
the equilibrium and motion of fluids. It has two 
divisions — Hydrostatics, which treats of fluids at rest; 
and Hydrodynamics, which treats of fluids in motion. 

Matter exists either in the solid, the liquid, or the 
gaseous state, although by sufficient changes of tem- 
perature and pressure any substance can be made to 
pass from one of these states to another. Solids have 
both volume and form and offer marked resistance to 
any change of either; liquids have definite volume but 
take the form of whatever holds them; while a gas has 
neither definite volume nor form and expands until it 
fills the vessel which contains it. 

Liquids and gases have one distinguishing feature, 
which is absent from solids, and which causes them to 
be classed together as fluids, i. e. a freedom of motion 
of their particles among one another which causes them 
to yield to any dividing force and to change their shape 
under the slightest effort. 

A " perfect fluid " is defined as a substance which 
yields at once to any effort, offering no resistance what- 
ever to change of shape, but no such thing exists in 
nature. All fluids are more or less viscous and offer 
some resistance to the movement of their particles 
among each other, and so, too, all solids are found to 
be more or less plastic and to flow if acted upon with 
sufficient force. What distinguishes a plastic solid 



8 Hydeomechanics 

from a viscous fluid is that the former requires a cer- 
tain magnitude of stress to make it yield, while any 
stress, however small, will make a fluid yield, provided 
it be applied long enough. 

Even gases offer some resistance to change of shape 
at a flnite rate, but, when in equilibrium, gases, all 
common liquids, and even such viscous substances as 
honey and tar, behave like perfect fluids. Hence, 
although the science of hydromechanics is founded 
upon the consideration of a purely ideal substance, its 
hydrostatical results are found to be verifled in prac- 
tice, and it is only when fluids in motion are being 
considered that it becomes necessary to take account of 
their viscosity. 

(2) A " fluid,^^ then, is an ideal substance which has 
to a perfect degree the distinguishing feature of gases 
and liquids ; i. e. it is a substance incapable of resisting 
change of shape, and, therefore, incapable of experien- 
cing distorting or tangential stress. 

It follows directly from this definition that the pres- 
sure of a fluid at rest upon any surface must always 
be entirely a normal pressure. 

This may be illustrated by the immersion of a thin 
plate in water, when it will be found that there is no 
resistance to the beginning of its motion in any direc- 
tion in its own plane, though when actual motion takes 
place there is, of course, a certain amount of frictional 
resistance varying with the speed. 

(3) And, furthermore, it is a necessary consequence 
of our conception of the nature of fluids that the pres- 
sure at any point in a fluid at rest must be equal in 
all directions. In other words, if an indefinitely small 
plane area be imagined at any point in a fiuid in equi- 



An Elementaby Teeatise 9 

librmm, not only is the pressure on that area normal 
to it, bnt the amount of the pressure is exactly the 
same no matter how the area is turned about. 

For imagine a minute cube of a fluid at rest taken at 
any point within it, and, neglecting the effect of gravity, 
consider what would result from the least excess of 
pressure upon one of its sides. Evidently the fluid 
would yield and flow through the other five sides. A 
solid cube only requires for equilibrium that the pres- 
sures on opposite sides shall balance, but a fluid cube 
will not retain its shape unless the same normal pres- 
sure is exerted upon each side. I^ow this reasoning is 
independent of any assumption as to the angle at which 
the imaginary cube stands. Therefore, for equilibrium, 
the^ pressure must be equal in all directions at each 
point. 

In fact, the principle of sufficient reason justifies the 
conclusion that the pressure at a point in a fluid acted 
upon only by surface forces is the same in all directions, 
since there is no determining cause why it should be 
greater or less in one direction than in another. 

That it is legitimate to neglect the effect of gravity 
in reasoning as to the equality of the pressures around 
a point in a fluid will be apparent when it is considered 
that the weight of any volume of the fluid is propor- 
tional to the cube of one of its linear dimensions, while 
the pressure on its surface is proportional to the square 
of the same dimension. Thus, as the volume of fluid 
considered shrinks to a point, its weight vanishes in 
comparison with the pressure on its surface, and so it 
is equally true for fluids acted on by gravity as for 
those acted on only by surface forces, that at any point 
within them the pressure is equal in all directions. 



10 Hydkomechanics 

The equality of pressures in all directions about a 
point is illustrated by the fact that the pressure per 
unit area, called the " intensity of pressure," is the 
same on the bottom of a tank of water as it is on the 
side of the tank immediately adjoining the bottom. 
If an appreciable area be taken, the total pressure upon 
it, as will be hereafter shown, varies with the depth of 
its c'enter of gravity, but " pressure at a point " means 
intensity of pressure, or the pressure on an area divided 
by that area when the area is indefinitely diminished, 
and it is this pressure which is independent of direc- 
tion. Thus the pressure on a safety valve in a boiler 
is the same whether the plane of the valve face be 
horizontal, vertical or inclined. 

(4) In addition to their inability to withstand any 
tangential or shearing force, all fluids possess another 
important property. They have absolutely perfect 
elasticity of volume. That is, every change in the sur- 
face pressure acting upon a fluid causes a corresponding 
change in its volume and density. 

But from this it necessarily follows that at every 
point in a fluid subjected only to surface forces the 
pressure must be the same. For imagine two con- 
tiguous portions of such a fluid to be under unequal 
pressures ; then the part under the greater pressure will 
be of greater density than the part under the less pres- 
sure; but, if so, the former will expand and press 
against the latter until the forces of elasticity, lessening 
on the one side and increasing on the other, balance at 
the bounding surface, and both portions are equally 
compressed. 

(5) There is a marked distinction between gases and 
liquids in the matter of compressibility, the former 



An Elementaey Treatise 11 

being susceptible of great changes of volume as the 
surface pressure is altered, while the changes of volume 
of the latter are almost imperceptible even when the 
pressure varies from one to the other of attainable 
extremes. This is the cause of the common but fal- 
lacious opinion that liquids are incompressible, and 
for this reason they are sometimes distinguished from 
gases as being " incompressible ^^ or '^ inelastic ^^ fluids, 
gases being classified as " compressible " or " elastic " 
fluids. The fact, however, is, as already stated, that 
every fluid, whether a liquid or a gas, has perfect 
elasticity of volume; i. e. the change in volume per 
unit volume is proportional to any small change in 
pressure, whatever the pressure to which the fluid is 

subjected may be. In other words — = — kdp for all 

fluids, k being very small and constant, or nearly so, for 

liquids and being very nearly equal to - for gases. 

P 
Thus for example, the volume of atmospheric air would 
be reduced approximately yj by an increase of pressure 
of one pound per square inch, while the volume of 
water would only be reduced -g-o-gWc P^^^ ^J "^^^^ same 
increase of pressure. 

(6) It is this perfect elasticity of fluids which causes 
them to transmit, with undiminished intensity, to every 
part of their enclosing vessels, any surface pressure 
which is applied to them at any point. Of course, 
fluids upon the earth are subjected to the attraction of 
gravitation, and they may be similarly subjected to 
other attractive or repulsive forces which act upon the 
individual particles constituting their mass, and these 
so-called " bodily forces,^^ proceeding from external 



12 Hydkomechanics 

bodies^ may, and as will be seen do, produce in fluids 
pressures which vary from point to point, but omitting 
them from consideration for the moment, it is evident 
that there is no other way of producing pressure in a 
mass of fluid than by applying pressure at some point 
or points of an enclosing vessel, and it is such surface 
pressures which the elasticity of fluids causes to be 
transmitted to all points within them. 

Thus the atmospheric pressure, approximately 15 
pounds per square inch, is transmitted to every point of 
a liquid in an open vessel, so that if the liquid had no 
weight it would exert a pressure of 15 pounds upon 
every square inch of the vessel enclosing it. 

(7) The fact that if a fluid is acted upon only by 

surface forces the intensity of pressure is uniform all 

over its surface and at all points throughout its mass 

can also be derived from a 

/^^^\ T-^ ^ T consideration of the nor- 

yt_— _jF\^ - '* ^'i-* mality of fluid pressures. 

/^ ^=^ -^^^K For let Fig. 1 represent a 

/e - -^=---^ -^^^^y\ closed vessel full of a fluid 

t - !zr ~^j^^^zfr\ upon which gravity does 

py -^^^^:'?^^^^ ~^'Lz\ ^0^ act, and let the inten- 

t'.t^L^^^=^zS^:>J^ sity of pressure at A be p. 

E::i^.-£rr-^^?^5r^^r^ B Then, to prove that it is 

\^^r^.--=^;^^^^^ the same at every other 

point, take a right cylinder 

of the fluid standing on the surface at A as base, and 

of unit area cross-section, and suppose it to become 

solid. It will remain in equilibrium since nothing has 

occurred to alter the forces acting on it. But the 

forces acting on its sides, being pressures normal to its 

surface, have no components along the axis AP. 



An Elementaky Teeatise 13 

Hence the pressures at P and at A, each acting at 
right angles to the unit cross-section of the cylinder, 
must be equal, or the intensity of pressure at P is the 
same as at A. But since the pressure is the same in 
all directions at a point in a fluid, we can now suppose 
the unit area at P to be turned about so as to be parallel 
to any other part of the containing vessel, say at B, 
and then by taking a cylinder of fluid from P to B we 
see that for equilibrium the pressure on its ends must 
be equal, and thus the pressure intensity at B is the 
same as at P, and so on. Hence the pressure is the 
same at every point in the fluid, and, if we apply a 
certain pressure at A by means of a piston of one 
square inch area, we thereby produce that same pres- 
sure on every square inch of the surface, so that a 
piston of 100 square inches area at B could only be 
held stationary by a force 100 times as great as is 
applied at A. 

This is the principle of the hydraulic press, and by 
means of it force can be multiplied almost indefinitely, 
always, of course, at the expense of speed. Thus, in 
the above example, if 100 pounds is applied to the 
small piston, the large one will exert a pressure of 
10,000 pounds, but the work done by pushing the 
former in only equals that done by the outward motion 
of the latter, a movement of one inch of the small only 
causing one of .01 of an inch on the part of the large 
piston (neglecting compression of the fluid, which, in 
the case of a liquid, may be done without sensible error). 

If gravity is acting, the weight of the particles of 
^uid produce, as will be shown, a pressure increasing 
with the depth, but this pressure is merely superim- 
posed upon that produced by the surface forces, and 



14 Hydkomechanics 

in no way affects the principle of the transmission of 
the surface pressures in all directions with undimin- 
ished intensity. 

(8) It has already been pointed out that all fluids 
have perfect elasticity of volume, and that the two 
classes of fluids, liquids and gases, differ greatly in 
their compressibility. 

A " perfect gas '' is defined as one whose compressi- 
bility of volume is numerically equal to the intensity 

of its pressure. In other words, for a perfect gas, — ^ 

V 

= p, or — = £ , whence by integration, pv = 

constant, or, if Vq is the volume under pressure po , 
pv = k = poVo . Now, while there are no gases in 
nature which exactly obey this law, which is known 
as Boyle's, or as Mariottes', law, it is found experi- 
mentally that all gases come very near to obeying it, 
except when near liquefaction, and it may be considered 
to be practically true. 

In the equation pv = k, the constant k is only so 
when the temperature of the gas is constant, other- 
wise, by the law of Charles, or of Gay-Lussac, k = 
ko (1 + at), where a = -^\^ when t is expressed in 
centigrade degrees. Hence, combining both laws, 

STTT-t = m^" "'' *=^"'"s T (= ^'^ + *) '""'^ 

absolute temperature, we have for the general equation 
of the changes in volume, pressure, and temperature of 

any given mass of any gas, ^ = constant; and this 

equation is practically true for all gases except when 
they are near the point of liquefaction. 



An Elementary Treatise 15 

(9) In considering the effect of snrface pressures 
upon liquids, unless the utmost refinement of results is 
desired, they may be considered incompressible, and 
consequently of unchanging volume and temperature, 
but, in the case of gases, considerable changes, both of 
volume and of temperature, may result from a change 
in the surface pressure. Thus the hydraulic press, 
while enabling us to multiply force at will, transmits 
energy, or work, from one point to another without 
sensible loss, but if it were filled with a gas instead of 
a liquid, the work done upon the gas, compressing it 
and raising its temperature, would be an important 
factor. 

Here we will consider only the effect of surface pres- 
sures upon gases in the two extreme cases; first, of 
isothermal, and second, of adiabatic transformation. 

In the first case the temperature of the gas is sup- 
posed to remain unchanged, or at least to be the same 
at the end as at the beginning of any change of pres- 
sure, and, consequently, we have, under such circum- 
stances, the relation pv = k = p^Vo , or the volume 
varies inversely as the pressure. If we double the 
pressure upon any given mass of gas, we half its vol- 
ume, etc., and this is true provided the pressure is not 
so great as to have brought the gas near liquefaction, 
and provided further that the operation is performed 
in such a way that there is no change in the tempera- 
ture of the gas. 

In the second case we assume that the gas neither 
gains heat from, nor loses it to, other bodies, and in 
this case pv^ = k' = ip^Y^y (where r is the ratio of the 
specific heat at constant pressure to that at constant 
volume for the gas in question, its value being 1.408 
for air and practically the same for all gases), and this 



16 Hydkomechanics 

again is true provided the gas is not near liquefaction 
and provided further that the change in volume takes 
place either in a non-conducting envelope or so rapidly 
that the gas has not time to either absorb or give out a 
perceptible quantity of heat. 

In both cases the intensity of pressure is the same 
at each instant all over the surface and everywhere 
throughout the mass of the gas, neglecting increase of 
pressure with depth caused by the weight of the gas 
itself, which is usually immaterial. 

For example, if we have a tube 10 inches long, full 
of air at 15° C, and at atmospheric pressure (15 
pounds per square inch), and if we compress it to a 
length of 1 inch so slowly that the temperature remains 
unchanged, the final pressure will be 150 pounds per 
square inch. If, on the other hand, we compress the 
same air from 10 inches to 1 inch suddenly, the final 

1.408 

pressure will be 15i[-i.'>y= 15 X 10 ^ 383.8 pounds 

per square inch, and at the same time the temperature 
will be raised to 458° C, the latter value being given 

by the formula ^ = Mi^ whence T^ = (273 + 15) 

J- Po^o 

^^=737, orti = 458°C. 

(10) That the surface of separation of two fluids 
which do not mix possesses properties peculiar to itself 
is illustrated by the fact that a needle may be made 
to float, and certain insects can walk upon the surface 
of water. This and other so-called capillary actions 
are caused by molecular forces, which produce a 
certain surface tension, uniform in all directions, but 
depending for its amount upon the nature of the fluids 
in contact. It is this surface tension which makes 



An Elementaey Treatise 17 

fluids sometimes behave as if they were enclosed in 
an elastic skin, as when, for example, a drop of mer- 
cury on a table assumes a nearly spherical form, the 
effect of gravity, which depends upon the volume, 
being then small as compared with the tension of the 
surface. The surface tension, however, differs from 
that which would result from an plastic skin in not 
changing in intensity with expansion or contraction of 
the surface, but remaining always the same per unit 
length for a particular pair of fluids. Experiment has 
shown the value of the surface tension, in grains per 
linear inch of the surface, to be 1.26 for petroleum, 
3.23 for water, and 21.53 for mercury, each being in 
contact with air. 

It will be seen that it requires the expenditure of 
energy to extend the surface of a fluid, and that in con- 
tracting its surface a fluid does work, and since, calling 
the surface tension T, it will require T units of work to 
increase by unity the length of a strip of fluid surface 
of unit width, every fluid possesses an intrinsic energy 
T per unit area of its surface. 

Problems I. 

(1) If p is the intensity of a uniform fluid pressure, 
the units being the pound and the inch, what is the 
total pressure on an area of 2 square feet ? 

(2) A fluid exerts a uniform pressure on a surface of 
one square yard. If the total pressure is 5000 pounds, 
what is the pressure per square inch ? 

(3) It is desired to test a boiler by an internal pres- 
sure of 625 pounds per square inch. What force must 
be applied to a piston of one inch diameter to produce 
this ? 

2 



18 Hydromechanics 

(4) If the steam pressure in a boiler is 10 atmos- 
pheres, taking the atmospheric pressure at 15 pounds 
per square inch, what weight must be used to load a 
circular safety valve of 2J inches diameter to prevent 
escape of steam ? 

(5) A closed vessel filled with fluid has a number of 
pistons fitted in ijts walls. If one piston, having an 
area of 10 square inches, is pushed inward with a force 
of 40 pounds, what change results in the total pressure 
upon another piston of one square foot area? 

(6) If the centers of two circular pistons, respectively 
of 4 and of 12 inches diameter, are at the same level, 
and the smaller one is pushed inwards with a force of 
100 pounds, how much force must be applied to the 
larger one to keep it from moving ? 

(7) If two circular pistons in the walls of a vessel 
full of fluid have diameters respectively 3 inches and 18 
inches, what pressure on the smaller will produce a 
pressure of 1800 jDOunds on the larger? 

(8) The lever of a hydro-static press is 3 feet long ; 
the distance from its fulcrum to where it acts on the 
small piston is 8 inches; the diameters of the pistons 
are respectively one foot and -J an inch. If a pound is 
applied to the end of the lever, what weight will the 
large piston just lift? 

(9) What steam pressure in a boiler will just lift its 
safety valve if the ratio of its arms is 18 to 2 and a 
weight of 25 pounds is hung on the long arm, the 
diameter of the valve face being one inch? 

(10) If the pressure in the accumulator of a hydraulic 
forging press is 2 tons per square inch, and the diameter 
of the press cylinder is 80 inches, what is the power 
of the press ? 



Ax Elemextaky Teeatise 19 

(11) The energy of recoil of a 13-mch gun is 280 
foot-tons, and it is stopped in 4 feet by a 13-inch 
hydraulic cylinder whose piston is pushed into the 
cylinder by an 8-inch rod, the resistance being caused 
by forcing water from the pressure to the exhaust side 
through openings designed to produce constant pres- 
sure. If the water displaced by the piston rod escapes 
through spring valves loaded to 800 pounds per square 
inch, what is the pressure on the recoil side during 
recoil ? 

(13) If 100 cubic inches of air at 15 pounds per 
square inch pressure be raised from a temperature of 
20° C. to 50° C, what will be the volume if the pressure 
is kept constant, and what the pressure if the volume 
is kept constant? 

(13) If the density of the air at the earth's surface is 
4 times as great as at a height of 7 miles, and if the 
temperatures are 22° C. and 0° C, what would the 
volume of a balloon which held 640 cubic yards of 
hydrogen gas at the surface be when it had risen 7 
miles, provided its envelope offered no resistance to 
expansion ? 

(14) If a certain mass of air at 15° C. and 15 pounds 
per square inch pressure is compressed adiabatically 
till its temperature is raised to 110° C, what is its 
pressure and how does its volume compare with its 
original volume ? 

(15) What is the pressure of atmospheric air which 
has been slowly compressed to -^ of its normal volume ? 
and what if it has been very rapidly compressed to the 
same volume? 

(16) If a mass of fluid is falling freely in a vacuum, 
what shape will it assume, and why ? 



20 Hydromechanics 

(17) The 5-meter Whitehead torpedo has an air flask 
of 11.17 cubic feet capacity, and is charged to a pres- 
sure of 1500 pounds per square inch. If atmospheric 
air at 14.7 pounds pressure weighs .076 pounds per 
cubic foot, what is the weight of the air in the flask 
and what would its free volume be ? 

(18) If rain drops O'.'l in diameter are each formed 
by the coalescence of one thousand million smaller 
drops, how many foot-pounds of energy would be set 
free if 100 cubic feet of water were formed into drops ? 



An" Elementary Treatise 21 



CHAPTEE 11. 

Fluids Under Gravity. 

(11) Having seen that the pressure is everywhere the 
same within and over the entire bounding surface of a 
fluid acted upon only by surface forces, we will now 
consider the effect of the force of gravitation upon 
fluids, that being the only bodily force which acts upon 
them under ordinary conditions. 

In the first place, it is clear that the free surface of 
a fluid in equilibrium must be horizontal, or at right 
angles to the direction of the force of gravity. For 
suppose it is not, but at some point is inclined; then 
the force acting upon a particle of the fluid at that 
point can be resolved into rectangular components one 
of which is parallel to the surface, and since this latter 
force would be a tangential, or shearing force, the fluid 
could offer no resistance to it, and so the particle would 
move; and such unbalanced forces, preventing equi- 
librium, must exist until the surface everywhere is at 
right angles to the lines of action of the forces of 
gravity. The same reasoning, of course, shows that the 
free surface of a fluid can only be in equilibrium when 
it is normal at every point to the resultant of all the 
forces acting on it at that point. 

A gas cannot ordinarily have a free surface, but 
must be kept in a closed vessel to prevent its bemg 
diffused, but in any fluid at rest under the action of 
gravity, whether a liquid or a gas, the pressure is the 
same at all points in the same horizontal plane. For 



22 Hydeomechanics 

suppose A and B are two such points, and imagine a 
right cylinder of the fluid, of unit cross section, and 
whose axis is the line joining A and B, to be solidified. 
Then, since none of the forces acting are in any way 
changed, it will remain in equilibrium. But the pres- 
sures on its curved surface, as well as the forces of gravi- 
tation acting upon its particles, are all at right angles 
to its axis, and so have no components along that axis. 
Consequently the fluid pressures on its ends must them- 
selves balance and be equal, or the pressures at A and 
B are equal, and so for all points on the same level 
within the fluid. 

(12) Now consider two close horizontal planes within 
any fluid at rest, and let p be the pressure at the upper 
level, p + dp that at the lower level, and dy the 
distance between them. Imagine a rectangular parallelo- 
piped, of unit cross section, and whose upper and 
lower bases are in these two levels, to be solidified. 
It will remain at rest, since the forces acting upon it 
are unchanged. But the pressures on its opposite ver- 
tical sides are normal to those sides and so have no 
vertical components. Therefore its weight must be 
balanced by an excess of pressure on its lower over that 
on its upper base, or, if w is the weight of unit volume 
of the fluid at the level under consideration, p-f- dp = 
p + "^dy, or dp ^ wdy. If now w be variable, we 
have p = wdy, in which w must be expressed in terms 

of the depth before integration becomes possible, but, 
if the fluid be homogeneous, so that w is a constant, 
we have 

(1) p =: wy 4- po 

where y is reckoned downward from any assumed level 
and Po is the pressure at that level. 



Ax Eleme:^taey Treatise 23 

Thus it is seen that in a homogeneous fluid the effect 
of gravity is to produce an increase of pressure directly 
proportional to increase of depth. If gravity did not 
act, the pressure would everywhere be equal to that 
existing at the surface, — ^where there was a free surface 
it would equal the pressure of the atmosphere, — but 
since gravity always does act, the pressure actually is 
that at any point of the surface plus w}^, where w is 
the weight of unit volume of the fluid and y is the 
vertical depth below the selected point on the surface. 
This is only exactly true when the density is uniform 
throughout the fluid mass, which, since all fluids dimin- 
ish in volume and so increase in density as the pressure 
upon them increases, is never the case. But liquids 
are so nearly incompressible, and gases are so light, 
that the change in their densities due to their ovni 
weights may be neglected excepting when very great 
depths are being considered. Practically there are 
only two such extreme cases — the flrst when the pres- 
sure due to the atmosphere is to be determined, and 
the second when the effects of the pressure at great 
depths in the ocean are to be calculated. 

(13) The effect of gravitation upon a liquid, then, 
is to produce at every point within it a pressure de- 
pending solely upon its density and the depth of the 
point below the free surface, and to this must be added 
any existing surface pressure to give the total pressure 
at the point. Thus the pressure due to gravity at a 
point 100 feet below the surface of a lake is 100 x 62.5 
pounds per square foot (fresh water weighs 62.5 pounds 
per cubic foot), or 43.4 pounds per square inch, and to 
get the actual pressure at that depth the atmospheric 
pressure on the surface, 15 pounds per square inch. 



24 



Hydromechanics 



must be added to this, giving 58.4 pounds per square 
inch. 

In the case of a gas, where there is no free surface, 
gravitation has exactly the same effect, and the pressure 
at any point equals the sum of the surface pressure and 
that due to the depth of the point. Thus in Fig. 2, 

representing a closed 
vessel full of a gas, the 
pressure at the level 
AB equals the surface 
pressure at C plus the 
weight of a column of 
gas hi high, or, what 
is exactly the same 
thing, it equals the sur- 
face pressure at D plus 
the weight of a column 
hg high, the surface 
pressures at C and D differing solely by the weight of a 
column h2 — h^ high. 

So, too, in a liquid, the pressure at any level equals 
that at any other selected level plus that due to the 
weight of a column of the liquid whose height is the 
difference of the two levels, but it is most convenient 
to take the free surface of a liquid, if, as is usually the 
case, there be one, as the upper level, and then the 
pressure at any point is the sum of the pressure on 
that free surface (usually the atmospheric pressure) 
and that due to the depth of the point below the free 
surface, which depth is called the " head " at the point. 
(14) It follows that areas of equal pressure in all 
fluids at rest under gravity are horizontal planes, and 
that all parts of the free surface of a liquid must be 




An Elementaey Treatise 25 

at the same level however much the continuity of the 
surface be interrupted by solid bodies. Yery large 
water surfaces, of course, have a spherical contour, 
being, when at rest, everywhere normal to the direc- 
tion in which gravity acts. Variations in the atmos- 
pheric pressure, however, may produce considerable 
differences of level at widely separated points. 

(15) In the solution of problems it is frequently 
convenient and useful to apply the principle that areas 
of equal pressure are level surfaces by regarding even 
homogeneous fluids as being constituted of layers, each 
of which produces a surface pressure which is trans- 
mitted with undiminished intensity to all points below 
it. Thus in considering the pressure on any body 
which is sunk below the surface of a fluid, we can 
treat it as if just immersed and then merely add to the 
results obtained on this supposition the effect of the 
uniform intensity of pressure produced by the layer of 
fluid which stands above the highest point of the body. 

It is of the utmost importance, when numerical 
results are sought, to express every factor in the units 
of one particular system and in no others. Thus in 
calculating the pressure p due to a depth y in a fluid 
of which unit volume weighs w, by the formula p =^ 
"^y + Poj> if '^ is in pounds per cubic foot, then y 
should be expressed in feet and po in pounds per square 
foot, when p will be found in pounds per square foot. 
Whereas, if w is the weight of a cubic inch, y should 
be expressed in inches, and po in pounds per square 
inch, when the resulting value of p will also be in 
pounds per square inch. 

(16) We will now show that the common surface of 
two liquids which do not mix must be level. For, 



26 



Hydromechanics 




Fig. 3. 



referring to Fig. 3, take any 
two points, A and B, at the 
. same distance Z below the 
upper surface. Then, since 
the upper surface has been 
proved to be necessarily level, 
the two points A atid B are 
on a level, and therefore the 
pressure at each must be the 
same, say p. Now, taking 
vertical lines at A and B, let 
x' and x" be the distances on them from the upper to 
the common surface, and let w' and w" be the weights 
per unit volume of the upper and lower liquids respec- 
tively. Then at A we have p == x'w' + (z — x')w", 
and at B we have p ^ x"w' -\- {z — x")w"; whence, 
subtracting one from the other, x'(w' — w") — 
x"(w' — w") = 0. Therefore, since by hypothesis 
w' — w" is not zero, x' — x" must be, and the common 
surface, being thus equally distant at every point from 
the upper surface, must itself be level. 

(17) A distinction has thus far been made between 
liquids and gases by considering that the former alone 
have free surfaces, but there is one case, that of the 
earth's atmosphere, where a gas has a free surface, and 
where, too, the height of that surface above the earth 
is so great as to make it necessary to take account of 
the variation in density due to gravitation. 

If the density of the air were uniform, the total 
height of the atmosphere would be given by the for- 



mula p = Hwq , or H : 



in which H is called the 



' height of the homogeneous atmosphere,'^ and po and 



Ax Elemextaey Treatise 27 

Wq are respectively the pressure and the weight of unit 
volume of air at sea level; and, since po is about 14.7 
pounds per square inch and the weight of a cubic foot 
of air at 0° C. is .0807 pounds, this gives H = 

^^'\^}^^ = 26.^00 feet (about). 

Actually, however, the density of the air diminishes 
as we ascend, on account of the decreasing pressure, 
and so a vastly greater height of the atmosphere is 
requisite to cause the pressure which is found to exist 
at the earth^s surface. Moreover, the density of the 
air is affected by its temperature, and as this also de- 
creases with increasing height, we can only approxi- 
mately solve the problems of determining the true 
pressure at any given height and the true height of 
the total atmosphere. 

(18) Let Wq be the weight of a cubic foot of air and 
Po the pressure per square foot, at sea level, and let 
w and p be the corresponding values at the height y. 
Then dp =3r — wdy ; but, assuming the temperature to bo 



constant, 


w 

Wo 


= 


S-' 


whence dp = 


Wo 

~ p^ 


pdy, 


dp 


Wo 


dy 


= 


— 


i *■■ 


the integration of ^ 


rtdiich 


gives 


(3) 


lOge 


p 


— log 


.Po=-^ 


, or p 


= Po< 


_ y 

^ H 



From the first of (2), if p^ and pg are the pressures 
at heights y^ and j^ > we get by subtraction y^ — yi = 

TT 

— (log 10 Pi — log 10 P2)^ where fj- is the modulus of 
the common system of logarithms. But pressures are 

XT 

proportional to barometer readings, and — = about 



28 Hydeomechanics 

60,300 feet = about 10,000 fathoms, so we can write 
(3) J2 — Ji = 10,000 (log bi — log bs) fathoms, 

and we have the simple rule that the difference in 
height between two places, in fathoms, is 10,000 times 
the difference of the common logarithms of the simul- 
taneous barometer readings at the two places. 

Of course an exact formula for determining heights 
by the difference of barometer readings requires a 
number of corrections to be applied, the most im- 
portant being for difference of temperature between the 
two places, for the variation in the force of gravity due 
to height, and for the effect of aqueous vapor in the 
air on the pressures. Our value of H is based upon 
the assumption that the atmosphere consists of pure 
dry air at a uniform temperature of 0° C, but for 
exact results the value of H corresponding to the actual 
average conditions between the two places at which 
observations are taken must be used. 

Again, putting (2) in the form log^ ^ = _ ^ y, 

Po Po 

and changing to common logarithms, we have 
(4) y = — 10,000 log^ fathoms. 

Po 

Thus, to find the depth below sea-level at which the 
pressure is two atmospheres, we have y = — 10,000 
log 2 = — 3010 fathoms = about 18,000 feet. 

y 

(19) From the formula p = p^e ^ we see that an 
infinite height is necessary to give p a zero value, which 
would indicate that there is no limit to the height of our 
atmosphere. But this is doubtless due to the erroneous 
assumption of a uniform temperature. In fact an ex- 
treme upper limit to the height of the atmosphere is 



Ax Elementaey Teeatise 29 

found to be about 5J times the radius of the earth, 
because at that height the centrifugal force due to the 
earth^s rotation would balance the attraction of gravi- 
tation and so the particles of air, if there were any 
there, would fly off into space. 

If the laws which govern the fall of temperature 
with increasing altitude were known, and its effect 
allowed for, it is probable that the atmosphere would 
be found to extend at most but a few hundred miles 
above the earth^s surface. It is at least -certain that 
our formula assuming a uniform temperature errs on 
the side of overestimating the pressure at any given 
height, and as by that formula the height corresponding 
to a pressure of only .01 pounds per square foot is 
about 60 miles, it can be seen how almost infinitely 
tenuous the air becomes at such heights. 

The lowest actual barometric record ever made was 
in the balloon ascent of Glaisher and Coxwell in 1862, 
when, at an estimated height of 7 miles the barometer 
stood at 7 inches and the thermometer at — 12° F., 
indicating a pressure of about 3J pounds per square 
inch and a density of the air less than one-fourth that 
at sea-level. 

Problems II. 

(1) What is the pressure on an area of one square 
inch 40 feet below the surface of a lake ? What on a 
square foot ? 

(2) What depth of fresh water, and what of sea 
water would cause a pressure equivalent to that of the 
atmosphere (weights per cubic foot respectively 62^ and 
64 pounds and atmospheric pressure 14.7 pounds per 
square inch) ? 



30 Hydromechanics 

(3) What is the intensity of the pressure 1000 
fathoms deep in the sea? 

(4) Prove that^ if two fluids which do not mix meet 
in a bent tube, the heights of their upper surfaces above 
their common surface are inversely as their densities. 

(5) If a cubic inch of mercury weighs half a pound, 
what is the pressure on the bottom of a vessel 5 inches 
deep, full of mercury, and having a bottom of 1 square 
foot area? 

(G) A cubical tank one yard on a side is full of water 
and has a pipe leading into it which is itself full of 
water to a height of 10 feet above the surface of the 
water in the tank. What weight must be put on the 
top of the tank to keep it closed? 

(7) What is the pressure per square inch at the bot- 
tom of a vessel which contains water to a depth of 6 
inches, and oil 10 inches deep on top of that, if the oil 
weighs 0.9 as much as the water per unit of volume ? 

(8) A vessel 8 inches high contains sea water to a 
depth of 5 inches, olive oil one inch deep on that, and 
alcohol above that to the top. What is the intensity 
of pressure at the bottom, the ratios of the respective 
weights of the three fluids to that of an equal volume 
of fresh water being 1.027, 0.915 and 0.795? 

(9) Equal volumes of oil and alcohol are poured into 
a circular tube so as to fill half the circle. Show that 
the common surface rests at a point whose angular 
distance from the lowest point is tan~^ /y? 

(10) If two fluids are put in a circular tube so that 
each occupies 90°, and the diameter joining the two 
open surfaces is inclined at 60° to the vertical, find 
the ratio of the densities of the fluids. 

(11) If the average depth of the ocean is 12,000 feet. 



An Elementary Treatise 31 

and water is reduced one thirtieth in volnme for ever}^ 
10^,000 pounds pressure per square inch, how much 
would the surface rise if gravitation ceased to act? 

(12) Given the volume of an air hubhle 40' below 
the surface of a lake, where the temperature is 4.4° C, 
what will it be when it rises to the surface and has a 
temperature of 10° C. ? 

(13) A circular hollow cone with open base, of weight 
enough to sink, is suspended with axis vertical and 
vertex at a given depth below the surface of the water; 
what is the volume of the air compressed in the cone 
if its temperature remains unchanged ? 

(14) What would be the volume of the compressed 
air in the preceding case were the cone lowered to its 
immersed position very suddenly ? 

(15) A cylindrical diving bell of height a is sunk in 
water till the water rises half way up its axis. Show 

that the depth of its top is h — -t , h being the height of 

a water barometer, and show that, if the temperature 
of the air be raised t° C, the water will recede approxi- 

niately %^r^ 
^ 4 h + a* 

(16) The barometer reading at the foot of a moun- 
tain is 29.75 and at its top is 22.80; what is the differ- 
ence of level ? 

(17) How deep must a mine be that the pressure of 
the atmosphere at its bottom may be 2^ times that at 
its mouth ? 

(18) At what depth in water would the air in a bub- 
ble equal in density the surrounding water, given that 
water has 800 times the density of atmospheric air ? 

(19) If an air bubble has a volume of 3 cubic inches 



32 Hydromechanics 

at a depth of 10' in water, at what depth will its volume 
be 2 cubic inches ? 

(20) A balloon half full of coal gas just floats in the 
air when the barometer stands at 30; what will happen 
if the barometer falls to 28? What would happen if 
the balloon had been full at the first pressure? 

(21) A weightless piston fits into a vertical cylinder, 
of height h and cross section A, closed at its base and 
full of atmospheric air. The piston being originally 
at the top of the cylinder water is poured slowly on 
top of it. How much can be jDOured in before it will 
run over? 

(22) Equal quantities of fluids, of densities p and (t, 
fill half a small uniform circular tube in a vertical 
plane. Show that the radius to the common surface 

makes with the vertical tan~^ ^ ^ . 

p -\- <r 

(23) A circular tube of small uniform bore is half 
filled with equal volumes of four liquids which do not 
mix and whose densities are in the ratios 1:4:8:7. 
Show that the diameter joining the free surfaces makes 
tan~^ 2 with the vertical. 

(24) A cylinder is filled with equal volumes of n 
different liquids which do not mix. If the density of 
the upper liquid is p ; that of the next 2 /> ; of the next 
3/?; and so on; prove that the mean pressure on the 
corresponding portions of the curved surface are as 
1^ : 22 : 32. .. : n^. 



An Elementary Treatise 33 



CHAPTEE III. 

Total Pressures — Centers of Pressure of Plane Areas. 

(20) Having learned that the pressure at any point 
in a fluid is independent of direction, that it is normal 
to any surface in contact with a fluid, and that its 
intensity is the sum of the surface pressure and a 
pressure due to the weight of a column of the fluid 
of height equal to the depth of the point below the 
surface, it is next in order to determine the amount of 
the pressure on any given immersed surface and its 
line of action. 

Let S be any surface, whether plane or curved, and, 
taking axes of X and Y in any horizontal plane and 
the Z axis vertical, let po be the pressure at the level 
XY and let p be the pressure at any point of the 
immersed surface. Then p = p^, -j- wz, where w is 
the weight of unit volume of the fluid, and, if ds be 
an element of the surface, the whole pressure on the 

surface is P= pds, the integration extending over the 

whole surface. Substituting the value of p, we have 

^ ^^ (Po + ^z)ds = poS + w zds. But zds = zs, 

where z is the distance of the center of gravity of the 
surface from the plane of XY. Hence we have 
(5) P = S(po + wz). 

In other words the total pressure is the s^ne as if the 
whole surface were concentrated at its center of gravity. 



34 Hydeomechanics 

In the case of liquids, having a free surface, and 
neglecting the atmospheric pressure, this formula may 
be stated in words as follows: The whole pressure on 
any immersed surface equals the weight of a column of 
the liquid whose cross-section equals the area of the 
surface and whose height is the depth of the center of 
gravity of the surface below the surface of the liquid. 
And this rule applies to all surfaces, whether plain, 
curved or warped, and however they may be situated 
with respect to the surface of the fluid. 

Thus, for example, the total pressure on the surface 
of a sphere of radius a just immersed in a liquid is 
4ra^ X wa = 4-a^w, the center of gravity of a spherical 
surface being, of course, at its center. Again, the 
center of gravity of a zone being midway of its altitude, 

the total pressure on the upper hemisphere is 2r a^ X^y- 
= n a^w, while that on the lower one is 2 tt a^ X -J^ = 

Srra^w, the sum of these being the total pressure on 
the sphere as before. 

(21) It is evident, too, that in the case of a vessel 
containing fluid, the whole pressure on the surface is 
exactly the same in amount as if the vessel were im- 
mersed in fluid to the same depth, the only difference 
being that in the first case the pressure is from in out 
and in the second case from out in; this, of course, on 
the supposition that the walls of the vessel have no 
thickness. For, if the immersed vessel be filled to the 
level of the outside fluid surface, there must be com- 
plete equilibrium, since by supposition the vessel is 
merely a geometrical locus, as a spherical surface for 
example, and so the inside pressures must just balance 
the outside and vice versa. 



An" Elemektaey Tkeatise 



35 



(22) It is important to note, however, that this total 
pressure on a curved surface is merely the sum of all 
the pressures acting on the elements of the surface, 
which, unless the surface be a plane, have various direc- 
tions, and there is no physical reality which corresponds 
to such a sum. It is only the total pressure on a plane 
surface, or the sum of the components in some fixed 
direction of the pressures on a curved surface, which 
has a distinct physical meaning, being the total force 
with which the surface is urged to move in that 
direction. 

Eor example, let us determine the sum of the vertical 
components of the pressures on a sphere just immersed 
in a fluid. Fig. 4 
represents a vertical ' 

section of the sphere, 
whose radius is a, 
and the equation of 
this section referred 
to axes as shown is 
x' + y' == a^. If 
now P be any point 
on this circle, its 
depth below the sur- 
face is a — J (y being 
negative for the point P in the figure), and the pressure 
is the same at all points on the small circle of the 
sphere whose diameter is PP'. Therefore, if we take 
an element of the vertical circle, ds = ad^, the total 
pressure on a strip of the sphere of that width, and at 
the level of P, is 2-xds X (a — y)w = 27rawx(a — y)d^ 
= 2n-a% sin^ (1 + cos ^) d 0. But, since every element 
which goes to make up this total pressure is normal to 




--:-.P 



TlQ.A. 



36 Hydromechanics 

the sphere, each of them makes the angle e with the 
vertical, and so the vertical component of the total 
pressure on the strip is Srra'w sin 6 cos 6(1 + cos <?)d e, 
and the total upward pressure on the sphere is 

27rav[sin e cos 0{l + cos ^)d ^ = 2 ra^-wf^^ _ ^^J' 

_ ^a w ^ Which, as of course it should be, is the 
o 

weight of the fluid contents of the sphere. 

The total pressure on the sphere has been shown 

to be 47ra^w, or three times as much as the upward 

pressure just found, but the former is nothing but the 

sum of the normal pressures acting all over the surface 

of the sphere, and is incapable of being represented by 

any single resultant, while the latter is the sum of the 

forces which urge the immersed sj^here upwards, and 

can be represented by a single resultant force acting, 

as will be hereafter seen, through the center of the 

sphere. 

(23) It must be remembered that it is the depth of 
the center of gravity of the surface, not of the solid 
it may enclose, which determines the total pressure. 
Thus the whole pressure on the curved surface of a 
cone of radius of base a, height h, and vertical angle 
2a^ immersed with axis vertical and vertex at the sur- 

face, is^^^i?^ X -~, the center of gravity of 

the surface being at two-thirds the altitude from the 
vertex. 

(24) The next point to consider, having learned how 
to determine the amount of the pressure on an im- 
mersed surface, is at what point does it act. In other 
words at what point of a surface can a single force 



An Elementaky Tkeatise 37 

equal to the total pressure on the surface act so as 
to completely replace the system of forces which are 
the pressures on the various elements of the surface. 

N"ow in general a system of non-coplanar forces can 
only be replaced by a single force when the said forces 
are parallel. It is true that the fluid pressures on any 
closed surface have a single resultant which acts 
through the center of gravity of the enclosed volume, 
but this special case will be considered hereafter. For 
the moment we will only consider the case of plane 
areas under fluid pressure^, and the problem of deter- 
mining the line of action of the resultant of the pres- 
sures on such surfaces amounts simply to finding where 
the resultant of a system of parallel forces acts, since 
all the fluid pressures are normal to the surface and 
therefore parallel. 

(35) The point of action of the resultant of the fluid 
pressures on a plane area is called the '^ center of pres- 
sure " of the area. If the area is horizontal, its center 
of pressure is evidently at its center of gravity. Other- 
wise it must always be below that point, since the fluid 
pressure increases with the depth. 

Suppose F(xy) :^ to be the equation to the curve 
enclosing the given area, the axis of x being the inter- 
section of the plane of the area with the surface of the 
liquid, and a being the angle that plane makes with a 
horizontal plane. Then, if p is the pressure at any 
point whose coordinates are x, y, p = wy sin a (ne- 
glecting surface pressure); the pressure on an element 
of the area is pdxdy; the moment of that pressure 
about the axis of x is pydxdy; and the sum of the 

moments of aU the fluid pressures is pydxdy = 



38 



Hydromechanics 



w sin affyMxdy. Hence, if y be the ordinate of the 

center of pressure, and P be the total pressure on the 
area, we have 

(6) yP =: w sin a||y2dxdy 

And similarly, taking moments about the axis of Y, 
we have 

(7) xP = w sin ot j I xydxdy. 

Usually the total pressure P is most readily found by 
multiplying the area (A) by wh, where h is the depth 
of the center of gravity of the area, but when either 
A or h is not already kno^n, we have 

(8) P = w sin aj j ydxdy, 

the integration, of course, extending in all cases over 
the whole area under consideration. 

It will be observed that sin « cancels out in the 
determination of x and y, showing that the position of 
the center of pressure in the plane of the area is inde- 
pendent of the angular position of the plane, and this 

is as it should be, 
since swinging the 
plane about its in- 
tersection with the 
liquid surface makes 
exactly the same 
proportional change 
in the pressure upon, 
every element as it 
does in the total 
pressure. 




FiO. 



An Elementaey Treatise 



39 



(26) It is often unnecessary to resort to double inte- 
gration to find centers of pressure, since an elementary 
strip, all parts of which are at an equal depth, and 
upon which the pressure is consequently uniform, 
can usually be selected, and the total pressure upon it, 
and its moment, be at once written down. For ex- 
ample, take the case of a segment of a parabola im- 
mersed vertically with vertex at the surface, as shown 

in Fig. 5, the equation to the curve being x" = J . 

If now we take an elementary strip, area 2xdy, the 
intensity of the pressure all over it will be wy, and so 
the whole pressure on it will be 2wxydy, and the 
moment of that pressure about the axis of x will be 

fa 2wb r^ 

2wxyMy = -^ yidy = 
o 3i^ Jo 



4wa^b 



But P 



2w 



4wa^b 



5a 



Therefore y = — - and, the curve being symmetrical to 

the axis of Y, the center of pressure is on that axis 
and x = 0. 

As another exam- 
ple, take a triangle 
of base b and alti- • 
tude a, immersed as 
shown in Fig. 6. Let 
X be the width of the 
triangle at the dis- 
. tance y from its ver- 
tex. Then the pres- 
sure on the elemen- 
tary strip xdy is 




40 



Hydeomechanics 



wyxdy and its moment about X is wy^xdy, and the sum 

of the moments of all the pressure on the triangle is 

wb fa ,, / . X y \ wa'b -r, < 
since ^ = -^ ] = — —. ±>ut 



I wy'xdy = ^~ J yMy ( 



a 
2aw 



y \ _ wa'b 
"a y — "IT" 



„ ab 



wa^b Tx - wa^b wa'b 3a 
^. Hencey = — ^^=^. 

Sometimes polar coordinates lend themselves better 
than rectangular to the determination of centers 3f 
pressure. Thus to find the center of pressure of a 
circular disc of radius a vertically immersed with its 
upper edge tangent to the surface. Taking the origin 
at the surface, which is almost invariably advisable, 

the equation to the 
circle is r = 2a cos 0. 
Then taking rd dr 
for the element of 
area, the intensity of 
the pressure upon it 
is wr cos 6, since 
r cos is its depth 
below the surface; 
the total pressure 
upon it is wr^ cos^ 
pressure about x 
so we have yP = 

571 

cos" d = 4wa' X 




_6-a*w 



But we also have P= 



n c2a COS e 

wr^cos d drdo -. 



16 

8wa' 
'■ 3 



Jcos* ^ d ^ = 

Sra^w __ 5a 
4:7ra^w ~ 4 ' 



8wa^ 
3 



dr: 



i^w. Therefore y = 
The value of P, the total pressure on the 



An Elementaey Treatise 



41 



circle, might, of course, have been written down 
directly, without the use of the double integral, since 
it is merely the weight of a column of the fluid whose 
cross-section equals the area of the circle and whose 
height is the depth of the center of gravity of the 
circle. 

(27) In the calcu.lations of centers of pressure thus 
far made the areas have been taken as just immersed, 
for the reason that this is the simplest case, but, of 
course, the position of a center of pressure varies with 
the depth, and it is frequently necessary to determine 
it when the area is sunk below the surface. This can, 
however, be easily done by considering that the fluid 
above the highest point of the area plays the part of an 
atmosphere, causing a uniform intensity of pressure at 
every point below it. It is only necessary, then, to 
combine the resultant of the variable pressures due to 
the fluid actually in contact with an area with the 
resultant of the uniform pressure due to the fluid above 
the area, and this will give the total pressure and the 
center of pressure of the area in its sunk position. 

Thus, for example, if a vertical circle of radius a is 
immersed with its highest point c below the surface, 
as in Fig. 8, the uni- 
form intensity of pres- 
sure due to the super- 
imposed fluid would be 
cw, and the resultant 
would be a pressure 
Tra^cw acting at the 
center of gravity of 
the circle. But the 
pressure due to being 




Fig. 8. 



42 



Hydromechanics 



just immersed is tt a^w and acts at a distance - below 
the center of the circle. Therefore the total pressure 
on the circle is w-a^(a + c) and it acts at a distance 

'^ below the center, that being, therefore, the 

4(a + c) 

position of the center of pressure of a circle placed with 
its highest point c below the surface of a fluid. 

As a rule the atmospheric pressure acts upon both 
sides of the surfaces whose centers of pressure have to 
be determined, and so can be left out of account, but 
in case one side of a surface immersed in liquid has no 
fluid pressure whatever upon it, the uniform pressure 
upon the other side due to the atmospheric pressure on 
the liquid surface must be combined with the pressure 
of the liquid itself in the same way as has just been 
explained. 

(28) Among practical problems requiring knowledge 
of centers of pressure is that relating to the stability 
of brick or masonry embankments against one or both 
sides of which water stands. For example, we will de- 
termine how high water can stand on one side of a brick 
wall 12 feet high and 3 feet thick, the weight of the 
wall being 112 pounds per cubic foot. Let h be 

the height of the water 
when just about to push 
the wall over. Then the 
pressure P on each foot 
length of the wall is 

:^ ^ X^^ = 31.25h2; and 

this pressure acts -^ above 

Yl G. 9. ^^® ^^^® ^^ ^^^ wall, since 




An" Elementary Treatise 



43 



the center of pressure of a rectangle just immersed is 
f its altitude below the surface. ISTow the wall may 
give way either by sliding horizontally or by turning 
about its outer lower edge at A, and in practice failure 
will first occur by the latter method. Therefore, when 
the wall is just about to upset, the moment about A 
of the weight of the wall must just equal the moment 
about the same edge of P, which is the only upsetting 



'2 



X 36 X 112, 



^ hP 3W 31.25h^ 

lorce. Hence -— = — — ; or — - — 

o /i o 

12 , 

or h = — V 42 z= 8.34 feet. 

(29) Another practical application is to determine 
the forces acting upon the supports of floodgates, dry- 
dock caissons, and similar structures. Take, for ex- 
ample, an automatic 
tide gate (Fig. 10) 
for draining a salt 
marsh, its action be- 
ing to open when- 
ever the head of 
water on the marsh 
side exceeds that on 
the ocean side, and 
to close when this 
condition is revers- 
ed by the rising of 
the tide. The pressure on the ocean side is P^ = 

wbhi X -^' ~ — «-- , applied -- above the sill (b being 
2 2 o 

the width of the gate), while that on the marsh side is 
Pa = — ^r^ applied -,^ above the sill. The resultant 




Fig. 10. 



44 Hydromechanics 

pressure, therefore, is P^ — 1*2=-^ (W — V)? and 

the height above the sill of its point of application is 
found, by taking moments about the sill, to be z = 
h, P, - h,P, h,^ - h/ m. . +1. • 

YcK^^rp^ = SiW-WY ^^' ^'^'^ ^^'''' '' '""P" 

ported by the hinge and the sill and carries the load 
Pi — p2 at the distance z above the sill, so that if R 
is the thrust on the hinges and a the height of the gate, 
again taking moments about the sill, we have Ra == 

(Pi — P2)z; whence R = ^ (K' - K)- 

Peoblems III. 

(1) A cubical vessel is filled with a homogeneous 
liquid. Compare the total pressure on its bottom and 
sides. 

(2) What is the total pressure on a solid right cone, 
radius of base 3' and height 4', suspended in water with 
axis vertical and vertex 10' below the surface? 

(3) A closed hemisphere, full of a fluid, stands with 
its base vertical; compare the total pressures on its 
plane and curved surfaces, and show that the sum of 
the components, perpendicular to the base, of the 
pressure on the curved surface equals the total pressure 
on the plane surface. 

(4) A triangle is immersed in liquid with its base in 
the surface; where must a horizontal line be drawn 
across it so that the total pressure on the two parts 
shall be equal ? 

(5) A parallelogram is immersed with one side in 
the surface ; how must a line be drawn from one end of 



An Elemextaet Teeatise 45 

that side so as to divide the area into two parts, the 
total pressures on which shall be equal? 

(6) A vessel contains mercury with a layer of water 
10" deep standing above it. A rectangular area 8" 
high is immersed vertically with its base parallel to and 
below the common surface of the two liquids. What 
must its position be so that the pressure of the mercury 
on the lower shall equal that of the water on the upper 
part of the rectangle ? 

(7) A cube full of liquid is suspended by a string 
tied to one corner. Compare the pressures on its faces. 

(8) What is the total pressure in tons on a triangular 
area of 100 square feet placed with its vertices 4, 8 and 
12 feet respectively below the surface of the water ? 

(9) A vertical water gate 40' wide has fresh water 
standing 20' deep on one side; how deep must salt water 
stand on the other side that the pressures on the two 
sides may be equal ? 

(10) ABCD is a parallelogram whose diagonals AC, 
BD, intersect at E, and AB is in the surface of a fluid. 
Show that the pressures on the triangles AEB, BEC, 
CED are as 1 : 3 : 5. 

(11) A hollow sphere is half filled with mercury and 
then filled up with water; compare the total pressures 
on the upper and lower hemispheres. 

(12) A rectangular sluice gate 12' square is placed 
vertically in water, and the pressure on the half cut off 
by a horizontal line is y^- greater than that on the half 
bounded by a diagonal. How high is the water above 
the top of the gate ? 

(13) A hollow right cone standing base down on a 
horizontal plane is completely filled with water. Show 
that thq weight of the cone must equal twice that of 



46 Hydeomechanics 

the water in order to prevent the cone from rising, 
and thence deduce the whole normal pressure on the 
curved surface and the consequent position of the 
center of gravity of that surface. 

(14) A mill race of triangular cross-section (base 
horizontal, vertex down) is closed by a sluice gate which 
is supported at the three corners of the triangle. What 
part of the total pressure on the gate is supported at 
each corner? 

(15) Find the center of pressure of a segment of a 
parabola immersed with its base in the surface of a 
fluid. 

(IG) What is the total pressure on the hinges of an 
automatic water gate 4' high by 5' wide when the sea 
stands 3' above the sill on one side and the marsh 
water V above the sill on the other? 

(17) A vertical rectangular masonry dam is 4' thick 
and weighs 140 pounds per cubic foot. Determine its 
height that the water pressure against one side of it 
may make it fail by sliding, the coefficient of friction 
being 0.75, and the height at which it will fail by 
rotation. 

(18) A dam of triangular cross section and vertical 
back is 3' wide at the base and 15' high. How high 
can the water stand before the dam will fail (a) by 
sliding? (b) by rotation)? the material weighing 140 
pounds per cubic foot and the coefficient of friction 
being 0.75. 

(19) The center of a vertical circle of 2' radius is 
10' below the surface; where is its center of pressure? 

(20) Where is the center of pressure of a quadrant 
of a circle immersed vertically with one edge in the 
surface ? 



An" ElemeisTtaey Treatise 47 

(21) Find the center of pressure of a rectangle ver- 
tically immersed, with one angle in the surface and one 
diagonal horizontal. 

(22) A circular sluice gate is to be hung on a hori- 
zontal axis so placed that the gate shall automatically 
close when water stands at a given height above its 
center. How far below the center- of the gate must 
the axis be put ? 

(23) Prove that the pressures on the curved surface 
of a cylinder full of fluid held with its axis making an 
angle l^ with the horizontal have a single resultant 
which makes the angle /5 with the vertical. 

(24) A hollow cone without weight is filled with 
mercury and then inverted so as to stand with its open 
base on a smooth horizontal table. What is the maxi- 
mum height of the cone which will cause it to remain 
full? 

(25) A heavy conical cup stands vertex up on a 
smooth level surface and water is gradually poured into 
it through a hole in its top. If the weight of the 
cup is f that of the water which would fill it, how high 
will the water be when the cup is on the point of 
rising ? 

(26) A cone full of water is suspended by a cord 
attached to a point on the rim. of its base. Prove that 
the total pressure on its curved surface and that on its 
base are in the ratio 1 -{- 11 sin^ « : sin' a, where a is 
the semivertical ansrle of the cone. 



48 Hydeomechanics 



CHAPTEE IV. 



Resultant Pressure on Curved Surfaces — ^Density and 
Specific Gravity. 

(30) We have seen how to determine the sum of all 
the fluid pressures on any immersed surface, whether 
plane or curved, and also, in the case of plane surfaces, 
how to determine the line of action of the single force 
which is the resultant of those pressures. 

In the case of curved surfaces, as already pointed 
out, the sum of the pressures bears no necessary rela- 
tion to their resultant, and in order to determine the 
latter it is usually necessary to first find its compon- 
ents in the vertical and in two horizontal directions. 

Take any surface whatever immersed in a liquid, and 
project it upon the level free surface of the liquid. 
Imagine now that the liquid contained within the ver- 
tical lines whose intersections with the liquid surface 
form the hounding curve of the projection is solidified. 
Then the only force which is opposed to the weight of 
the solidified mass is the vertical resultant of the fluid 
pressures upon the immersed surface which now forms 
the base of the solid. Therefore the resultant vertical 
pressure on any immersed surface equals the weight of 
the fluid which stands upon the surface, and acts 
through the center of gravity of that fluid. 

Next project the immersed surface upon any vertical 
plane and imagine the liquid within the bounding, 
horizontal, projecting lines to be solidified. Evidently, 
for equilibrium, the horizontal resultant, in a direction 



An Elementaet Treatise 49 

perpendicular to the assumed plane of projection, of 
the pressures on the immersed surface must be equal 
and opposite to the resultant of the pressures on the 
projection of that surface. Therefore the resultant in 
a given horizontal direction of the pressures on any 
immersed surface equals the whole pressure upon the 
projection of the surface on a plane perpendicular to 
that direction, and acts at the center of pressure of that 
projection. 

Hence, in general, to determine the resultant of all 
the fluid pressures on a curved surface, we must find 
the vertical pressure on the surface, and the horizontal 
pressure on it in two directions at right angles to each 
other; and then the resultant of these three forces is 
the whole resultant fluid pressure on the surface. 

This resultant will usually be a dyname, or the com- 
bination of a couple and a force which is not parallel 
to the plane of the couple, since, as a general rule, a 
system of forces, such as the fluid pressures on a curved 
or warped surface, which are neither coplanar nor 
parallel, cannot be balanced by any single force. 

(31) There is one case, however, in which the fluid 
pressures always have a single force as their resultant, 
and that is when the immersed surface is a completely 
closed one, so that it forms the boundary of a geo- 
metrical solid. For, if such a surface be both sur- 
rounded and filled with a fluid, it will be in equilibrium, 
being itself without weight, and if then its contents be 
solidifled, without change of weight or volume, it will 
remain in equilibrium, and, consequently, the resultant 
of all the fluid pressures upon it must be a force equal 
and opposite to the weight of its contents. In fact, 
the projection of a closed surface upon any plane, 
4 



50 



Hydromechanics 



when algebraically considered, is zero, which shows 
that there is no resultant horizontal pressure on such a 
surface, and the weight of the fluid which stands upon 
a closed surface is really the difference of two columns, 
one representing the upward and the other the down- 
ward pressure, their difference, which is the fluid con- 
tents of the surface, being the resultant vertical pres- 
sure. 

(32) The resultant, then, of the pressures on any 
completely surface immersed in a fluid is a single force, 
equal to the weight of the fluid which the surface will 
contain, and acting vertically upwards through the 
center of gravity of the volume enclosed by the surface. 

When, therefore, any curved surface can be com- 
pletely closed by the addition to it of plane surfaces, 
we can readily determine the resultant of fluid pres- 
sures on the curved part, for the foregoing principle 
gives at once the resultant of the pressures over the 
whole closed surface, and since the pressures on each 
plane face have a single force as a resultant, we have 
only to combine the latter, reversed in direction, with 
the former and the result will be the pressure on the 
curved part of the surface. 

Thus, for example, take a right cone full of a fluid 
and held with, its 



Fig. 1 1.. 




axis 

horizontal, and let us 
determine the result-ant 
of the pressures on its 
curved surface. The re- 
sultant of the pressures 
over the entire closed sur- 
face is the weight of the 

rra'^h 

fluid, — ^w, acting at 



An Elemej^taet Teeatise 51 

its center of gravity,— from the base. The pressures 

on the base have for resultant the force TraV acting at 

the center of pressure of the base which is -—below its 
^ 4 

center. Consequently the pressures on the ' curved 

surface of the cone must have a single force as their 

-a^h 
resultant, that force being the resultant of ^^w 

o 

acting as shown in Fig. 11 and -aV reversed in direc- 
tion. 

(33) It is rarely necessary, for any practical purpose, 
to determine the resultant pressure upon curved sur- 
faces excepting in the case of bodies floating wholly or 
partly immersed in a fluid. But the importance of that 
case justifies a repetition and enlargement of the argu- 
ment from which it was shown that the resultant pres- 
sure upon a closed surface is equal and opposite to the 
weight of the fluid it would contain. 

If we imagine any portion of a fluid at rest to be 
solidified, without cliange of volume or weight, it will 
remain in equilibrium, and consequently the resultant 
of all the fluid pressures over its entire surface must 
be a force equal and opposite to the weight of the 
solidified body. Now if we put another body, of differ- 
ent weight, but of the same form, in place of the solidi- 
fied fluid, evidently the pressure on its surface will 
remain unchanged and so there will now be two forces 
acting, the weight of the new body acting at its center 
of gravity and the resultant of the surface pressures, 
wTiich is equal to the weight of the displaced fluid and 
acts at its center of gravity. These two forces, one 
acting upwards and the other, downwards, can only 



52 Hydkomechanics 

balance when they are eqnal and when their lines of 
action coincide. Exactly the same reasoning applies 
to the case of a hody only partially immersed in a fluid; 
there can only he equilihrimn when the resultant of the 
fluid pressures is equal and opposite to the weight of 
the hody, and;, at the same time, that resultant must he 
equal and opposite to the weight of the fluid which 
the hody displaces. 

(34) The resultant of the fluid pressures on a floating 
body is called the buoyant effort of the fluid, and its 
point of application, which we have just seen is the 
center of gravity of the displaced fluid, is called the 
center of buoyancy of the body. 

If, then, a body is either wholly or partially im- 
mersed in a fluid, its apparent weight is reduced by 
the weight of the fluid it displaces, and if it floats in 
equilibrium, its weight must be exactly equal to the 
weight of the fluid it displaces, and its center of gravity 
and center of buoyancy must lie in the same vertical 
line. 

If the fluid is homogeneous, the center of buoyancy 
will coincide with the center of form of the immersed 
part of the body. 

If, on the other hand, the fluid is heterogeneous, it 
must, when at rest, consist of horizontal strata, of 
density increasing with the depth, and the displaced 
fluid will consist of portions of the same strata, so 
that the buoyant effort will equal the sum of the 
weights of the different layers displaced, and the center 
of buoyancy will necessarily be below the center of 
form of the displaced fluid. 

Thus, when a body floats only partially immersed in 
water, the buoyant effort equals the sum of the weight 



An Elementaet Teeatise 53 

of the water displaced and that of the air displaced, 
and the center of buoyancy is at the center of gravity 
of the compound body formed by the displaced portions 
of both fluids. 

However, since the weight of air is only about -g-J-^y- 
that of water, — a cubic foot of fresh water weighing 
62-i pounds while a cubic foot of air weighs only about 
IJ oz., — it is only in very refined observations that 
it becomes necessary to take displaced air into con- 
sideration. 

Moreover, while it is true that all fluids under the 
influence of gravity are necessarily heterogeneous, and 
consist of horizontal layers of density increasing with 
the depth, yet the differences in density due to a 
moderate depth are so small as to be negligible, and 
practically the portions of fluids, whether gaseous or 
liquid, in which bodies are immersed can always be 
considered homogeneous. 

(35) It should be clearly understood that the buoyant 
effort of fluids is entirely due to the increase of pres- 
sure with depth caused by gravitation. If a uniform 
pressure be applied all over any closed surface it has 
no resultant; it is the varying pressures on immersed 
bodies which have the vertical resultant we call buoy- 
ancy. Thus, neglecting the increased density of the 
water due to depth, the buoyant effort of the fluid 
pressures on a body sunk a thousand fathoms below 
the surface of the sea is exactly the same as if it were 
just immersed, — the uniform intensity of pressure 
caused by the superimposed layer of water, when ap- 
plied over the whole surface of the body has a zero 
resultant, and so the resultant of all the pressures on 
the body is always just equal to the weight of the 
water it displaces. 



54 Hydeomechanics 

(36) Let us now apply the principle of buoyancy to 
the solution of questions relating to the density of 
bodies. By " density " we mean the mass of unit 
volume. This term is sometimes improperly used to 
denote the iveiglit of unit volume, but we shall call the 
latter " specific weight." Thus if p is the density of 
any substance, and w its specific weight, w == pg, 
where g is the acceleration of gravity. 

" Specific gravity " is relative density as compared 
with a standard substance, usually pure water at 4° C, 
and consequently it is proportional to weight per unit 
volume. Thus, if W and M are the weight and mass 
of any given volume of a substance, and W^ and M^ 
are the weight and mass of the same volume of water, 

W Mg M 

^ = TT— = ITT- , and the specific gravity of the sub- 
stance is the number resulting from dividing W by Wi- 

(37) It will be seen, therefore, that if a homo- 
geneous body floats in a fluid, its density is 
to the density of the fluid inversely as its 
whole volume is to the volume of the fluid it 
displaces. For, since the floating body is in equi- 
librium, its weight pgV^ must equal the buoyant effort 

P Vi 
of the fluid, />igVi ; whence — = -~ . Thus if a 

homogeneous body floats with f of its volume im- 
mersed in a fluid, its density must be f that of the 
fluid, and if the fluid be water at the standard tem- 
perature, the speciflc gravity of the body is 0.75 

Furthermore, since, as we have seen, the weight of 
any body immersed in a fluid is diminished by its 
buoA^ancy, which itself equals the weight of the fluid 
displaced, the specific gravity of any body must equal 



An- Elementaky Tkeatise 55 

its true weight divided by its loss of weight in water. 
Thus if a piece of steel weighing 28.33 pounds is im- 
mersed in water and then found to weigh 24.71 pounds, 
the difference of those weights, 3.62 pounds, is evi- 
dently the weight of the steeFs volume of water, and 
therefore the specific gravity of the steel is |^g| = 
7.826. 

Again, if a piece of cork weighing one pound in air 
is placed in water it will be found to require a force 
of about Si pounds to keep it wholly immersed, whence 

the specific gravity of cork is — = 0.24, since the 

1 + 'Je 

buoyancy, or weight of the cork^s volume of water, is 
evidently 3-J pounds more than the one pound weight 
of the cork itself. 

(38) If the specific gravity of a liquid be required, 
it is only necessary to take a solid of greater density 
than either the liquid or water, and to weigh it when 
immersed in each. Then, since its loss of weight in 
the two cases will be the respective weights of the 
same volume of the liquid and of water, the loss of 
weight in the liquid divided by the loss of weight in 
water will be the specific gravity of the former. 

It is a simple matter, too, to find the volume of a 
solid, however irregular its shape, by determining its 
loss of weight when immersed in a liquid of known 
density, such as water, and, if the solid is of less 
density than the liquid, the same problem can be 
solved by attaching to the solid a heavy substance 
which will sink it. Thus let V be the volume and W 
the weight of a solid, let w be the weight of unit 
volume, or " specific weight," of the liquid, and let 
X be the weight, when immersed in the liquid, of a piece 



56 Hydeomechanics 

of lead heavy enough to sink the solid. Then, if the 
solid, with the lead attached, weighs W^ when im- 
mersed in the liquid, we have x + W — wV = W^ , or 
^, ^ X + W - W, 
w 

(39) The balloon furnishes a good example of a body 
floating wholly immersed in a fluid, and in unstable 
equilibrium, rising when the weight of the air which 
it displaces exceeds its own weight, and falling when 
that condition is reversed. 

We will consider three cases, the first being the 
purely ideal one, in which the envelope of the balloon 
is without thickness or weight, perfectly elastic, and 
offering no resistance to stretching, and no weights are 
carried. 

If, then, Wq is the weight of the gas, which in this 
case is constant, and if at any moment the specific 
weights of the gas and of the surrounding air are 
respectively w' and w, the corresponding volume of 

Wo 

the balloon is Y = — 7 , and the weight of the air it 

w 
displaces is wV = Wq — 7 . Therefore the lifting force 

is F = Wof — , — 1 j. But, assuming gas and surround- 
ing air to be always at equal temperatures, since the 
pressures of the two fluids must always balance, their 
densities will vary exactly in the same proportion as 

w 

the balloon rises or falls, and so — , is a constant. 

w 

Hence, supposing the specific weight of the gas to be 

n times that of air, we have for the lifting force the 

constant value given by the equation — 



Aif Elemextaet Treatise 57 



(9) F = W„(^-1 

and such a balloon would rise with the constant 

Fg /I \ 
acceleration ™- = g f 11. Thus, for example^, a 

pound of hydrogen^ which will displace 14 pounds of 
air at the surface of the earthy will always displace 14 
pounds of air, however high it rises, provided merely 
that it can expand freely. 

(40) As the second case, we will suppose the en- 
velope to be of fixed internal volume Y^ , and that the 
total weight carried, including the envelope, is W^ in 
air. Then if the balloon is partially filled with the 
weight Wo of a gas of specific weight V, the volume 

Wo 

of air displaced by the gas will be —j , and the lifting 

force will be given by the equation — 
(10) F = W„^,-W„-W, 

w 

But in this case — - is only constant so long as the gas 

can expand freely in proportion with the decreasing 
pressure of the surrounding air: the moment the 
envelope is full of the expanded gas w' becomes con- 
stant, while w continues to decrease as the balloon 
rises. 

There will be a constant acceleration, therefore, 
until the envelope is fully expanded, after which the 
lifting force will be YqW — Wq — W^ . Consequently 

the balloon will cease to rise when w = — ^-^ , or, 

in other words, when the specific weight of the dis- 



58 Hydkomechanics 

placed air equals the total weight of the balloon 
divided by its capacity. 

(41) It will be observed that W^ was stated to be 
the " weight in air " of the solid parts, and so is 
really variable, increasing as the balloon ascends into 
less and less dense air, but the error resulting from 
assuming W^ to be constant is so small that it may 
be neglected. For, if W^ be 2000 pounds, supposing 
the average density of the materials to be the same 
as that of water, this volume would only be about 32 
cubic feet, displacing about 2J pounds of surface air, 
and so the increase of W^ , due to the balloon's rising 
to a great height, could hardly exceed the immaterial 
amount of 1^ pounds. 

Pkoblems IY. 

(1) The s. g. of mercury is 13,596, and 27.73 cu. in. 
of water weigh one pound; how much does a cubic 
inch of mercury weigh? 

(2) The s. g. of lead is 11.4; what volume of lead 
will weigh a ton ? 

(3) A piece of marble of s. g. 2.7 weighs 1000 
pounds; what is its volume? 

(4) A cubic foot of steel weighs 480 pounds, what 
will it weigh when immersed in water? What force 
would be required to keep it wholly immersed in 
mercury ? 

(5) A piece of lead weighs 47.48 grains in air and 
43.33 grains in water. What is its s. g. ? 

(6) A piece of platinum, weighing 2 pounds, weighs 
1.915 when suspended in olive oil, and 1.907 when 
suspended in water. What are the s. g.'s of the 
platinum and of the oil? 



An" Elementaey Tkeatise 59 

(7) A block of wood weighs 48 pounds in air; a 
piece of lead weighs ISJ pounds in water; when fas- 
tened together in water they weigh 4 pounds. W^at 
is the s. g. of the wood ? 

(8) The s. g. of cork is 0.34. What weight will 3 
cubic feet of cork sustain in sea water ? 

(9) A vessel full of water weighs 5 J oz.; a piece of 
platinum weighing 29^ oz. is put in it and it then 
weighs 33 oz. What is the s. g. of platinum? 

(10) The s. g. of pure gold and of copper are 19.3 
and 8.62 respectively. What is the s. g. of an alloy 
of 11 partrs gold to one part copper? 

(11) Atmospheric air consists of oxygen and nitrogen 
in the proportions of 21 to 79 by volume and of 23 
to 77 by weight. Compare the densities of the three 
gases. 

(12) A piece of gun metal weighs 1057.9 grains in 
air, and 934.8 grains in water; what proportions of 
copper and tin does it contain, if their respective s. g.'s 
are 8.788 and 7.291? 

(13) How many gallons of water must be mixed 
with 10 gallons of milk of s. g. 1.03 to make the s. g. 
of the mixture 1.01 ? 

(14) A cannon cast from bronze of s. g. 8.55 weighs 
546 pounds in air and 462 pounds in water. Show 
that there must be a flaw in it and find its volume. 

(15) How thick must a hollow copper globe of 
exterior radius a be in order to just float in vrater, if 
the s. g. of copper is 9.0 ? 

^(16) A block of ice, volume one cubic yard, floats 
with /^ of its volume above the surface, and a small 
piece of granite is seen to be imbedded in the ice. 



60 Hydeomechanics 

What is the size of the stone if the s. g. of granite is 
2.65 and that of ice 0.918? 

(17) A diamond ring weighs 69J grains in air and 
64J grains in water. The s. g. of the material of the 
ring being 16^ and that of diamond 3^, what is the 
weight of the latter? 

(18) The s. g. of silver being 10.5 and of copper 8.9, 
in what proportion by weight must they be combined 
to form a compound which shall weigh one-ninth 
more in air than in water? 

(19) A crown made of gold and silver loses -^^ of 
its weight in water, while equal weights of pure gold 
and of pure silver lose respectively y\ and ^2_ j ^hat 
are the proportions by weight and also by volume of 
the gold and silver in the crown? 

(20) A hollow cylinder full of a liquid is held with 
axis vertical, determine the resultant pressure on the 
curved surface on one side of a plane through the 
axis. 

(21) If the cylinder is held with axis horizontal, 
determine the resultant pressure on the lower half of 
the curved surface? 

(22) A right cone full of liquid is held with axis 
vertical and vertex down. Determine the resultant of 
the pressures on the curved surface on either side of 
a vertical plane through the axis. 

(23) If the cone is suspended freely from a point 
in the rim of its base, prove that the resultant pres- 
sures on the curved surface and on the base are in 

the ratio ^ if the vertical angle of the cone is 60°. 

(24) A spherical shell is full of liquid. Determine 
the resultant pressure on the curved surface cut off 
by a vertical plane through the center. 



"An Elementaet Teeatise 61 

(25) If the plane, instead of being yertical, is in- 
clined at a given angle, determine the resultant pres- 
sure on each half of the sphere^s surface. 

(26) A copper sphere, .01 inch thick and 40 feet 
in diameter, weighs 2200 pounds. It is charged with 
175 pounds of hydrogen, which weighs .0056 pounds 
per cubic foot at the surface atmospheric pressure 
of 14.75 pounds per square inch, while the air at the 
same pressure weighs .0807 pounds per cubic foot. 
With what acceleration will the sphere rise ? What is 
the pressure of the hydrogen at the beginning? At 
what height will it equal that of the surrounding air ? 
How high will the sphere rise ? What will the pressure 
be inside and outside of it then? 

(27) A right cone full of water rests on its side on 
a level surface. Find the resultant vertical and hori- 
zontal pressures on its curved surface. 

(28) A solid hemisphere is just immersed in liquid 
with its base inclined at tair'^2 to the surface. Find 
the resultant pressure on its curved surface. 

(29) A cone floats with axis horizontal in a liquid 

of twice its own density. Find the pressure on its 

base, and prove that, if is the inclination to the 

vertical of the resultant pressure on the curved surface, 

, ^ 4 tan a . . ,. . ,. , , 

tan = , where a is the semi-vertical angle 

of the cone. 



62 Hydromechanics 



CHAPTEE V. 

Floating Bodies Continued — Stability. 

(42) In the two cases considered in the last chapter 
the envelope of the balloon was supposed to he com- 
pletely closed, and in the second of those cases the 
envelope was assumed to he capable of withstanding 
the increasing excess of internal over external pressure 
which would follow the ascent of the balloon above 
the height at which the envelope was completely ex- 
panded. In actual practice it is necessary to maintain 
equilibrium between the internal and external pres- 
sures by having the neck, or lower end, of the balloon 
constantly open. With this arrangement, if the bal- 
loon is not completely filled with gas initially, the 
action at first is precisely the same as in the preceding 
case, but, as soon as such a height is reached that the 
expanded gas has filled the envelope, a new action 
begins, since further ascent causes an escape of gas at 
the neck. If the balloon would have held P pounds 
of the gas, but starts up containing only P' pounds, 
then the height at which it is just full is that at which 

P' 

the density of the air is -5- times its density at the 

earth. We will suppose the balloon to be completed 
filled with gas at the earth's surface, since otherwise 
it will only be necessary to first find the height at 
which it is just full, and then to take that as a new 
starting point. 

Let Wo , then, be the weight of gas which just fills 



Aj^ Elementary Teeatise 63 

the balloon^ of capacity Yq; let W^ be the weight in 
air of all the solid parts; let w and nw be the respective 
specific weights of air and of the gas at the height Z, 
and let W be the weight of gas remaining in the bal- 
loon at the same height. Then, since the open neck 
keeps the gas at eqnal pressure with the surrounding 
air, n is a constant, the densities of the two fluids 
changing in the same proportion as the balloon rises 
or falls. Moreover, VnWo = Wq and Vnw = W (Wq 
and nWo being the specific weights of air and of the 

gas at the earth); so that -^ — — ; which is merely 

'V Wo 

to say that the weight of a fi:s:ed volume of the gas 
is proportional to its density, which is itself propor- 
tional to the density of the surrounding air. We have, 
then, for the lifting force at height Z, F = V^w — 

W- W, =^- W~Wi=Wo^fi-l)- W,; or, 



n 



'oV n 



w p 
snice— = — , where p is the pressure at the height Z 

Wo Po 

and Po the pressure at the earth's surface. 

The balloon will rise until F = 0, or — = ^ 



" w,(i-i)' 

but we have alreadv seen in Chapter II that — = E ^ ; 

^ Po 

Z W 

so we have — ^^ — loge — -—" ; wheuce Z ~ H loge 

Ml-') 

Mi-') 

— ^^ 1, or using common logarithms, 

Wi 



64 Hydromechanics 

(12) Z = 10,000 log^l ^ - l) fathoms. 

As an application of these formnlge, let us take a 
balloon of 28,000 cubic feet capacity, filled with hydro- 
gen, which has a specific weight jV "t^^t of air. Then 
n = Jj , and taking Wq as .08 pounds per cubic foot, 

Wo = Tj^^ "^ ^^^ pounds. Therefore, the lifting 

force at starting is Wof "^ - ij = 160 X 13 = 2080 

pounds; and, if the weight carried was 1000 pounds, 
the balloon would rise to the height 10,000 log fj^^ = 
3180 fathoms = 19080 feet. 

(43) In practice the ascent is regulated either by 
throwing over ballast to lighten the load, or by letting 
gas escape by means of a valve in the top of the bal- 
loon to reduce the lifting force. The pressure of the 
gas in the upper part of the balloon slightly exceeds 
that of the external air at the same level, since the 
pressure of gas and air at the neck of the balloon 
are always equal, and the weight of the column of gas 
from neck to top of balloon is less than the weight 
of a parallel column of the outside air. There is also, 
of course, a continuous loss of gas by diffusion, which, 
to some extent, affects the results attained in practice 
with balloons. 

(44) Thus far we have chiefly considered wholly 
immersed bodies, but most practical questions relate 
to the buoyancy and stability of bodies which float 
only partially immersed. We will, therefore, now 
take up the case of a body floating partly immersed in 
a heavy liquid, and in its consideration we will neglect 



An Elementary Treatise 65 

altogether the displaced air as of no influence in 
comparison with the displaced liquid. 

The section of a floating body hj the plane of the 
fluid surface is called the " plane of flotation/^ the 
line joining the center of gravity of the body and its 
center of buoyancy is called the " axis of flotation '' ; 
and the depth to which the body is immersed is called 
the " depth of flotation " or the " draft." 

As already pointed out, the axis of flotation must 
be vertical when the floating body is in equilibrium, 
and it is also evident that, if both the body and the 
fluid are homogeneous, the center of buoyancy will 
coincide with the center of gravity of the immersed 
part of the body. 

When the form of a body is known, its depth of 
flotation in a given fluid will determine its weight, and, 
similarly, if its weight is known, the depth of its 
flotation can be found. Thus, if the length and beam 
at the water line of a battleship are 435 and 76 feet 
respectively, the mean draft 24 feet, and the coefiicient 
of flneness 0.66, she will displace 435 X ^6 X 24 X 
0.66 cubic feet of water, which, at 64 pounds per cubic 
foot, will weigh 14,500 tons, which, consequently, is 
the true weight of the ship and all that she contains. 
Or, reversing the problem, and having given the 
actual weight of a ship, we can calculate the number 
of cubic feet of water which she will displace, and 
then, from the form of her underwater body, determine 
her draft. 

As a simple example, we will determine the depth 

of flotation of a homogeneous right cone floating with 

vertex down. Let w be the specific weight of the 

fluid and w' that of the cone, and let V and h be the 

5 



66 Htdeomechanics 

volume and height of the latter, and x its depth of 

flotation. Then, since the volumes of similar solids 

are proportional to the cubes of their corresponding 

linear dimensions, the volume of the displaced fluid is 

x^V x^V 

-rj. Hence the buoyant effort is ~tt"^> ^^^ ^^^^ 

must equal the weight of the cone Yw'. From which 

- X' w' ' /W 

we have r, = ~ or x = hJ — . 
h' w ^^ w 

(45) A problem of practical importance is to deter- 
mine the relation between the displacement of a ship 
and the change in her draft which results from going 
from salt into fresh water, or vice-versa. 

Let D be the displacement in tons, A the water 
line area in square feet, and x the change of draft in 
feet. Then, since sea water weighs 64 and fresh water 
62J pounds per cubic foot, the displacement in cubic 
feet in salt water is 35D and in fresh water it will be 
6?5 ^ ^^-'-^- Hence the change in the displaced 
volume is 35D(^5 — 1), and this must equal xA; 
from which we have 21D = 25xA, or 

(46) If the center of gravity of a floating body is 
not in the same vertical with its center of buoyancy, 
the weight of the body and the buoyant effort of the 
fluid form a couple, which, unless some extraneous 
force is applied to balance it, must continue to rotate 
the body until it assumes a position in which the two 
opposing forces act in one and the same straight line. 
If, then, a floating body in equilibrium be displaced 
by turning it through a small angle, its position is 



An ELEMEisn^AKY Tkeatise 



67 



said to have been one of " stable ^' or of " unstable " 
equilibrinm, according as the couple resulting from 
its angular displacement tends to turn it back to its 
original position, or still further away from that 
position. If, however, a small angular displacement 
does not produce any couple, the centers of gravity 
and of buoyancy remaining in the same vertical, then 
the body is said to be in a position of "neutral" 
equilibrium. 

(47) As the practical questions in regard to stability 



FiG.12. 




deal with floating bodies which are symmetrical, when 
in their normal position of equilibrium, with respect 
to vertical longitudinal planes through their centers of 
gravity, only such bodies will be here considered. 

Fig. 12 represents a vertical transverse section, 
through the center of gravity G-, of a floating body 
which has been displaced through the angle 0, IF be- 
ing the new position of the water line, which when 
the body was in its normal state of equilibrium was 
LL'. Let V be the displaced volume of liquid, and 



68 Hydromechanics 

suppose the center of buoyancy, which before heeling 
was at B, is now at B'. Then the weight of the body, 
acting vertically downwards at G-, and the equal 
buoyant effort of the liquid acting upwards through 
the new center of buoyancy B', form a couple, whose 
arm is GZ, which tends to restore the body to its 
former position, or to heel it still further, according 
to whether the point M, the intersection of the new 
vertical through B' with the old one through B, is 
above or below G. In other words, the righting mo- 
ment is W X GZ = W X GM sin 6?, W being the 
weight of the body and GM being positive when M 
is above G and negative when M is below G. 

The point M is called the transverse metacenter of 
the body, and so we have the rule that a floating body 
is in stable, unstable, or neutral equilibrium, according 
as its metacenter is above, below, or coincident with 
its center of gravity. 

Now the positions of B and G depend entirely upon 
the form of the body and its distribution of weights, 
so that the distance BG can be determined in each 
case. Therefore, if we can determine BM, we have 
only to compare it with BG to ascertain whether or 
not the body is in stable equilibrium. 

The transfer of the center of gravity of the dis- 
placed liquid from B to B' has been accomplished by 
removing the wedge of which LOl is a section and 
inserting the wedge of which L'ol' is a section, and 
the volumes of these wedges, which are called respec- 
tively the " out " wedge and the " in " wedge, must 
be equal, since the total displacement remains un- 
changed. Therefore, if v is the volume of either of 



Ax Elementary Treatise 



69 




these wedges, and e and i rep- 
resent their centers of gravity. 
Fig. I 3 ^e must have V X BB' = 
V X ei. 

Xow let XLX' (Fig. 13) be 
half the water-line section, and 
let y be the half breadth at 
the distance x, along the lon- 
gitudinal axis xx', from the 
transverse section through G. 
Then, if be small, dv, an 
element of the volume of the 



wedge, is y(9 X 

whence we have v X eo 

BB' = V X ei 



^ X dx, and eo equals fy; 



and Y X 



'3 






now d be indefinitely diminished, —^ becomes t^, 

where ds is an element of the curve BB' (Fig. 12), and 
d<? is the angle between successive normals to that 



curve ; and BM = B'M 



-^^^- d^ = 3v { 



y^dx. Ee turning 



now to the one-half water-line plan (Fig. 13), we see 
that the moment of inertia of that area about its 

longitudinal axis is y'dxdy = -J y^dx, so 

J-lJo •'-I3 

that we have 



(14) 



BJI=v 



70 Hydromechanics 

where I is the moment of inertia of the whole water- 
line area about its longitudinal axis. 

(48) In order, then, to investigate the stability of 
any floating body we must calculate the moment of 
inertia of its water-line area and divide that by its 
volume of displacement, thus getting the distance 
from its center of buoyancy to its metacenter; then, 
subtracting from this result the distance of the center 
of gravity above the center of buoyancy, we get what 
is called the " metacentric height,^' or the distance of 
the metacenter above the center of gravity, and, if 
this is positive, the body is in stable equilibrium; if 
negative, in unstable equilibrium; and if zero, in 
neutral equilibrium. 

(49) If the body be inclined through an increasing 
angle, the center of buoyancy moves in a curve which 
is called the curve of buoyancy, and the locus of M, 
called the metacentric locus, is the evolute of that 
curve, M being the center of curvature of BB', and 
its vertical height above B being always given by the 

equation BM = ^ where I is the moment of inertia 

of the momentary water-line area; provided, however, 
that* that area continues to be symmetrical to the ver- 
tical longitudinal plane through BM. 

With ordinary ship shape forms, however, the point 
M remains nearly stationary for angles of heel not 
exceeding about 10°, and for determining the righting 
moment within that limiting angle, we can practically 
use the value of BM found for the body's normal posi- 
tion of equilibrium. 

(50) As an example, we will now determine the 
conditions for the stability of a solid cylinder floating 



An Elementary Treatise 71 

with axis vertical. Let r be the radius, h the height, 
and hi the depth of flotation. The moment of inertia 

of the water-line area, is -j- , since it is a circle with a 

diameter as axis; and V = Trr^h-^ . Therefore BM = 
■p2 h h 1 

-,,- ; and since BG =-7 ^ , the metacenter is above 

4h, 2 2 

1^2 h — hi 
the center of gravity as long as -r- > ^ , which, 

accordingly, is the condition for stable equilibrium. 

As another example, take a log of square cross sec- 
tion, and of specific gravity S, and let us determine 
the condition that it may float with one face horizontal. 
Then if a be one side of the square section, 1 the 
length of the log, and h the depth of flotation, h = 
Sa and V = ahl = Sa^l. Moreover, the water-line 

area being a rectangle, I = — -. Therefore BM =j^ 

Also BG- =^ -^ — = — (1 — S). Hence the meta- 

2 2 /& 

centric height is BM — BG = ^ ^ (1 — S). 

12b 2 

Putting this equal to zero, we have S^ — S + ^ = 0, 

or S = i ± --=: = .789 or .211. From which we 

2^3 

conclude that the specific gravity must either be less 
than .211 or greater than .789 in order that the meta- 
center may be above the center of buoyancy, or for 
stable equilibrium. 

Problems V. 

(1) A balloon, of capacity 680 cubic yards, starts up 
with a gross weight of 1150 pounds, including a charge 



72 Hydeomechanics 

of 400 pounds of coal gas, the specific weight of which 
is one-third that of the air. With what acceleration 
does it rise? How high is it when the gas just fills it? 
How much higher will it rise without throwing over 
any ballast? How much gas has then been lost? 

(2) A balloon of 8000 cubic feet capacity, and filled 
with hydrogen, the specific weight of which is -^ 
that of the air, has to carry a gross weight of 450 
pounds (exclusive of gas), including an observer with 
instruments. How much ballast must be added so 
that the balloon shall rise with an accelerating force 
of 10 pounds? How high will the balloon then rise? 
How high will it rise if all the ballast is thrown over? 

(3) A solid right cone floats in a liquid with vertex 
'Up; find the position of equilibrium. 

(4) A hemispherical vessel of given weight floats in 
a liquid with one-third its radius immersed. What 
weight must be put into it to sink it another third 
of a radius? 

(5) A cylindrical pontoon with hemispherical ends, 
25' long and 2' in diameter, floats half submerged in 
water. What weight will it support? 

(6) A cube floats in a liquid with one corner below 
the surface and three corners in the surface. Show 
that the s. g. of the liquid is 6 times that of the cube. 

(7) A uniform rod has a small quantity of mercury 
in one end of it and is found to float half immersed 
and at any angle with the vertical. Prove that the 
weight of the mercury equals that of the rod. 

(8) How deep will a paraboloid of revolution of 
height h sink in a fluid whose s. g. is n times its own, 
the axis being vertical and vertex up ? 

(9) A cylinder is placed with axis vertical in a 



An Elemen-taey Treatise 73 

liquid whose density varies as its depth. If the den- 
sity of the cylinder is the same as that of the liquid at 
a depth equal to half the height of the cylinder^, at 
what depth will the latter float? 

(10) A vessel going from salt into fresh water sinks 
%" , but after burning 50 tons of coal rises V . What 
was her displacement? 

(11) A 5000-ton ship, drawing 25', has to discharge 
300 tons of water ballast to cross a 34' bar into a river. 
She burns 50 tons of coal in going up the river into 
fresh water. How much is her draft then, and how 
much ballast will be needed to increase it by 1'? 

(12) A cylinder floats in a liquid with its axis in- 
clined to the vertical at an angle tan-^|, and with its 
upper base just above the surface. Show that the 
radius is ^ the height of the cylinder. 

(13) A hollow right cone, with closed base, made of 
uniform thin material, is found to float wholly im- 
mersed in a liquid in any position in which it is placed. 
Prove that half its vertical angle is sin~^ \. 

(14) A solid right cone floats in water with axis 
vertical and vertex down. If 27. is its vertical angle, 
prove that its speciflc gravity must be greater than 
(cos a )^ for stability. 

(15) A solid cylinder, one end of which is rounded 
off with a hemisphere, floats with the spherical surface 
partly immersed. Determine the maximum height of 
the cylinder consistent with stability. 

(16) A solid paraboloid, height a and radius of base 
b, floats with axis vertical and vertex down. Show 
that the height of the metacenter above the center of 
buoyancy equals half the latus rectum. 

(17) A cone whose vertical angle is 60° floats in 



74 Htdkomechanics 

water, axis vertical and vertex down. Prove that its 
metacenter is in the plane of flotation and that its 
s. g. must be greater than f J for stability. 

(18) A hemisphere floats flat side down. Prove that 
the distance from its base to its center of gravity 
and to its metacenter are in the ratio of the densities 
of the fluid and the solid. 

(19) A prism whose right section is an isosceles 
triangle of given vertical angle floats in water with its 
vertical angle down. Determine its specific gravity 
for stable equilibrium. 

(20) The right section of a prism is a right angled 
triangle. If it floats with the right angle down and 
opposite side horizontal, show that the metacenter is 
as much above as the center of buoyancy is below the 
plane of flotation. 

(21) Show that a homogeneous sphere will always 
float in neutral equilibrium, and determine the meta- 
centric height if it is so weighted that its center of 
gravity is one-fourth the radius from the center. 

(22) A cone of density p floats with an element 
vertical in a fluid of density ^, the base being just out 

of the fluid; show that if 2o.\'i the vertical angle, — = 

(cos a )2, and that the length of the vertical side 
immersed is to the length of the axis as cos 2 a : cos a. 

(23) A cone floats with axis vertical, vertex down, 

having — of its axis immersed. A weight equal to 

the weight of the cone is placed upon-the base and the 
cone sinks until just immersed before rising. Prove 
n^ -f n2 + n = 7. 



An Elementary Treatise 75 

CHAPTEE VI. 
Liquids in Uniform and in Steady Motion. 

(51) It has already been pointed out that no fluids 
in nature are " perfect fluids/^ or incapable of sus- 
taining any tangential stress, but that, on the con- 
trary, all possess to some extent a property, known as 
viscosity, whereby an internal resistance in the nature 
of friction opposes itself to their motion and con- 
stantly converts into heat some portion of their kinetic 
energy. 

In the study of fluids at rest viscosity may be 
neglected without any appreciable error, but in the 
case of fluids in motion its effects are sometimes very 
great, and must be taken account of. It is necessary, 
however, to any simple mathematical treatment of 
the subject of hydrokinetics to neglect viscosity, and 
we shall therefore confine our mathematical investiga- 
tions to "perfect fluids,^^ taking account of the vis- 
cosity of actual fluids, so far as is possible, by intro- 
ducing into our results numerical factors determined 
by experiment. Moreover, we shall, in the main, 
confine our attention to liquids in motion, and for 
simplicity of treatment we shall consider them as 
incompressible, which is so nearly the case that no 
errors of any importance can arise from the assump- 
tion. 

(52) It has been shown that the pressure intensity 
^t any point in a fluid at rest is equal in all directions, 
and this is also true in the case of fluids in motion. 
For, if we take a small cube of a fluid, by D'Alembert's 
principle the reversed effective forces and the impressed 



76 



Hydkomechanics 



forces, which act upon the volume of the cube, taken 
together with the pressures on the faces of the cube, 
constitute a system in statical equilibrium. Now, if 
the cube be diminished indefinitely, the effective and 
the impressed forces, which are proportional to the 
volume, vanish in comparison with the pressures, which 
are proportional to the area of a face of the cube, and 
so finally the cube must be in equilibrium under the 
pressures upon its faces. But this can only be, in the 
case of a fluid cube, when the pressure upon each 
face is the same. Hence, it follows from the essential 
property of fluids— their inability to resist tangential 
stress, — that, whether at rest or in motion, at any 
point within them the pressure intensity is equal in 
all directions. 

(53) A fluid is said to have uniform motion when* 
the velocity of each particle of it is uniform through- 
out its whole path, and we will first consider, as an 
example of this simplest case, a fluid rotating about 
a vertical axis. 

If an open vessel containing a fluid be given a uni- 
form rotation about a vertical axis, friction will soon 

cause the fluid to 
Rt* I rotate, like a rigid 

body, with the same 
angular velocity, 
and its free surface 
will assume a con- 
cave shape with its 
lowest point at its 
intersection with 
the axis of rota- 
tion. 




Fi G.14-. 



An Elementary Treatise 77 

Let Fig. 14 be a vertical section of such a vessel, OY 
being the axis of rotation and (o the nniform angular 
velocity. Then, taking a particle of the fluid, of mass 
m, in the free surface, its coordinates being x and y, 
the impressed forces acting on this particle are mg, 
its weight, acting vertically down, and mw% the 
centrifugal force, acting at right angles to the axis of 
rotation as shown; and the surface of the fluid must 
stand at right angles to the resultant E of these forces. 

Hence we have tan <p =—^ = ; My = — xdx; 

dx mg ' J •" g J 

w^X^ 

y = — — , the equation to a parabola; so that the sur- 

face is a paraboloid of revolution. 

Moreover, since the motion is uniform, the pressure 
at any point P in the rotating liquid depends only 
upon the surface pressure and the depth, so that p = 
Po + '^ X PD, where po is the surface pressure and 
w the weight of unit volume of the fluid, whence, if z 
is the depth below 0, we have 



(15) p = p„ + w(z+— J 



2g 

Indeed this last equation can be deduced directly from 
the conditions of rotation, and thence the equation to 
the free surface can be obtained in the form z = — 

-— . For take any point P in the fluid, at the distance 

r from the axis of rotation and z below the level 

through 0. Then the increase of pressure due to 

moving a distance dr further out on the same level is 

dp w 

-i. dr = -w^rdr; and that due to an increase of 

dr g 



78 Htdeomechanics 

depth dz is ^ dz =: wdz. But the pressure is a func- 
tion of only the two variables r and z. Hence dp = 
Y^ dr + -y- dz = — w^rdr + wdz; or^ integrating, p =^ 

w 
wz + ^^ io^T 4- C. But when z = o and r == o, we 

2g 

have p = Po so that C = p^ and as before, p = Po + 

/ w'r'\ 

w(z + 2g )• Moreover, since the pressure must be 

the same all over the free surface, we get the equation 

w*r2 
to that surface by putting p = p^j , whence z = — ~2Z~ > 

showing that it is a paraboloid of revolution with 
vertex at 0. 

(54) As another case of uniform motion take the 
buckets of an overshot water-wheel. Let ABD (Fig. 15) 

represent a vertical sec- 
tion of a bucket which is 
J^ P moving with uniform 
^ angular velocity oj about 

*Q, a horizontal axis perpen- 
dicular to the plane of 
w(gV the paper at 0. Let P 
be any point in the sec- 
tion of the water sur- 
face AD. Then if m is 
P 1 cr the mass of a particle of 

^ ^ *^-^ *^* the water at P, the only 

forces acting on it are mg vertically downwards and 
niw^ X OP acting outwards along OP. Eepresenting 
these forces by PE and PF respectively, their resultant 
is PG, which cuts the vertical through at C, and 




An- Elementaet Treatise 79 

from similar triangles we have 77-pi = — „ ,7 ^-p> , or, 
* OP mw^ X OP 

00 = — Y = constant. Therefore, since the water 

surface must always stand normal to the resultant of 
all the forces acting on it, each vertical section of the 
water surface is the arc of a circle whose center is at 

or 

the fixed distance — ^ ahove the axis of the wheel, and 

so the surface in. each bucket is a cylindrical one, 
having for its axis a horizontal parallel to the wheeFs 

g 
axis and at the distance -^ ahove it. 

(55) The next simplest and by far the most im- 
portant case to be considered is that of steady motion 
under the action of gravity. 

A fluid is said to be in steady motion when at each 
point throughout its mass the magnitude and direction 
of the velocity at that point remains unchanged. In 
steady motion the lines of motion coincide with the 
actual paths of the particles of the fluid, which are 
called " stream lines. ^^ If, then, we take a very small 
closed curve in a fluid in motion, and at each point of 

its contour drawn 

a stream line, we ^^^ '=^' ■=r\-^^ T^— -^ ^^ 
have what is called Z^. zmi^r. — y'^^ ~^^p^ T^zz l .- 
a " stream- tube.^^ — • ^^-~ 7 VS - fe^^ -— ~^-"-— ~ 

At any two points - — • ~/' y - ' — ^,' 

Pi and Po (Fig. 16) ZE^. J^^^^^^fT'^Z^'ZI 

in such a stream =- • p.^/^ -|-r- _^-- • ^-=r-_- 
tube, in a liquid of -rEr-~ rz^Jz-rlL V^Il"-r^IL -r^ 
density p, take nor- -^ _ — — -^^ :— : — .- ■P.^irT -- 
mal sections, and let t-^ -TT" 

their areas be (t and r 1 J ♦ 



80 Hydromechanics 

<?! respectively: let the velocities and pressures at the 
two points be v, p, and v^, p^ : let their respective 
heights above an}^ assumed datum line AB be z and z^ ; 
and suppose that the motion is from P to Pj . Then, 
since the liquid is incompressible, and there is no 
motion across the stream lines which form the bound- 
ing surface of the tube, a mass pYff enters the tube 
at P in each unit of time and an equal mass p'^^^^ 
leaves it at P^ : whence we have v<7 = ^i<^i- Now 
the mass entering at P arrives with a total energy 

iov^ [- + gz V this being the sum of its kinetic and 

potential energies, and moreover the work done on 
the tube at P in each unit of time is pv<r . Similarly 

the mass leaving at P^ carries off the energy /JV^^r^ ( ~ 

+ gZi), and the work PiViO-^ is done by the tube at P.^ 
while it escapes. But, since the motion is steady, and 
assuming that there are no losses by friction, the por- 
tion of the stream tube being considered neither gains 
nor loses energy. Hence we must have pvc- + p\(T 

i}^ + gz) = PiVi'^i + />Vi^i (y + gZi),or, dividing by 

gp\(T = gp^i^i, and writing w, the specific weight of 
the liquid, for gp^ 

P ^'^ Pi v/ 

(16) ^ -I ^z = i^+^+z, 

^ ^ w ^ 2g ^ w ^ 2g ^ ' 

In other words, since P and P^ are any two points 

p v^ 
along any stream line, — +97" + 2 is constant for 

each stream line, though its value may change in pass- 
ing from one stream line to another. 



An Elementary Treatise 



81 



(56) Evidently— is the height of a column of the 

liquid which will produce by its weight the static 

pressure p, and this is called the " pressure head/^ 

v' 
Similarly the quantity -^, being the height in falling 

from which the Telocity v would be acquired, is known 
as the " velocity head '' ; and z, the height above an 
assumed level of zero potential, is called the " poten- 
tial head/^ 

Thus the mathematical result just deduced, and 
which is known as Bernoulli's Theorem, amounts to 
saying that at every point along a stream line in any 
moving frictionless liquid, the sum of the pressure 
head, the velocity head, and the potential head, is con- 
stant. 

(57) It is important to clearly understand the dis- 
tinction between " hydrostatic " and " hydrodynamic " 
pressure. If a liquid be at rest the difference between 
the pressures at any two points within it depends solely 
upon the difference of the heads at those points, but 
the moment that motion occurs, the pressure at any 

point depends not only 
upon the head but also 
upon the velocity at 
that point, and, other 
things being equal, the 
greater the velocity 
the less the pressure 
and vice versa. Thus, if 
water stand at the level 
00' in a tank (Fig. 17), 
and if small open tubes 



il!p 

iill 


..a... 


^4-...c}. 


.9... 


=l£I 


; 


'J- 




ml 


r. a- 


I 




T^''z 


'r. 


c 




=^T 


L^ 


tow. 


^=^% 



A 



B 
Fi qXI. 



82 Hydromechanics 

(called piezometers) are inserted as shown in tlie out- 
let pipe ABC, then;, while the water is at rest, it will 
rise in each tube to the same level 00' as it has in the 
tank. But if the outlet be opened so that the water be- 
gins to flow, the levels in the tubes will sink to the 
points a, b and c, showing that the pressures at the 
three points A, B and C, on the same level, but in dif- 
ferent portions of the pipe, are no longer equal, but 
that the smaller the section of the pipe and the greater, 
consequently, the velocity of the water, the less is its 
pressure. The distances aa^ , bb^ and cc^ represent the 
respective velocity heads at A, B and C, and the dis- 
tances Aa, Bb and Cc are the corresponding pressure 
heads; the sum of the two heads at each point being 
constant and, if there are no losses by friction, equal 
to the original hydrostatic head. 

The outflow pipe being full, it is evident that the 
product of the cross section at any point and the 
velocity of the liquid at that point must be constant, 
so that the smaller the section the more rapid must 
be the motion. But unless the pressure at B were 
greater than it is at A, the velocity at A would not be 
retarded to that at B, and, unless the pressure at B 
were greater than it is at C, the accelerated flow at 
C would not be produced. If the pressure in a wide 
section and in a succeeding narrower section were 
equal, the elements of the liquid would not be acceler- 
ated; would not escape fast enough; and so would 
accumulate at the entrance to the narrower section 
until a sufficient augmentation of pressure was pro- 
duced. 

(58) We will now apply Bernoulli's theorem to the 
case of a vessel kept filled to a constant level with 



Ax Elemektaey Teeatise 83 

liquid which escapes through a small aperture in its 
walls. We do this by considering the escaping liquid 
as a whole to be a stream tube, one normal section of 
which is at the free surface of the liquid in the vessel 
and another normal section where the issuing jet 
attains the atmospheric pressure. Then, if the cross 
section of the vessel is large as compared with the 
orifice, we can neglect the velocity at the free surface, 
and since the pressure there is also the atmospheric 

pressure, we have at the free surface — ^ -|- z == C, and 

v^ p v^ 

in the let ^ — h ^" + z., ^ C; whence — — =^ z — z, or 
J 2g w ^ 2g 

(17) v^ = 2gh 
where v is the velocity of efflux of the liquid and h is 
the difference of levels of the orifice and the free 
surface, or the head on the orifice. 

This theorem, that the velocity of efflux through an 
orifice is the same as would be acquired in falling 
through a height equal to the liquid head on the orifice, 
was first formulated by Torricelli, and is the founda- 
tion upon which the science of hydraulics is largely 
built. It simply amounts to saying that if there were 
no losses by friction the energy acquired by a mass of 
liquid in falling would just suffice to raise it again to 
the same height from which it fell. When tested 
experimentally we find that the reversed velocity of 
efflux almost, but not quite, returns liquid to its 
original level, and we conclude that the difference is 
to be accounted for by the viscosity of the liquid. 

(59) The importance of the formula v = \^2gh will 
justify its deduction in a more direct manner than by 
the use of Bernoulli's theorem. 



84 Hydromechanics 

Let an opening of area k be made in the bottom of 

a vessel containing a liquid of density /', and at any 

time thereafter let K be the cross section of the vessel 

at the surface of the liquid^, h the height of that surface 

above the orifice, and v the velocity of the escaping 

liquid. Then, if in the time dt the surface falls dh, 

in the same interval the liquid mass K^odh must escape 

v' 
through the orifice, having the kinetic energy Kpdh — • 

But the work done is the descent of the liquid mass 

K/)(]hfrom the surface to the orifice, or the distance h. 

v^ 
Thus we have Kgohdh = K^odh — , or v^ = 2gh. 

The only assumption we have made is that all the 
work done in the vessel appears as kinetic energy in 
the issuing jet. It is entirely immaterial by what 
path the liquid moves within the vessel; the whole 
work done must be equivalent to the transfer of the 
surface layer to the orifice, and this must equal the 
energy of the escaping liquid, provided we can neglect 
losses of energy in overcoming friction and imparting 
velocity to the liquid remaining in the vessel, and 
experiments show that we can practically do this if 
the orifice is small in comparison with the cross sec- 
tion of the vessel. 

(60) It will be noted that in the first deduction of 
the formula v = \/2gh, v, by hypothesis, is the 
velocity at a section of the issuing jet throughout 
which the pressure is the same as at its surface, and 
such uniformity of pressure can only occur where all 
the stream lines are parallel and the assumed section 
a normal one to the stream tube. So also, in the 
second deduction of v = V^gh, v, being merely that 



An Elementary Treatise 85 

velocity which, will account for the kinetic energy of 
the jet, can only be considered the true velocity of 
and move in parallel lines. xA.ctnally, however, the 
efflux, if all parts of the Jet have the same velocity 
escaping liquid is made up of a great number of ele- 
mentary streams converging towards the orifice, and 
so the motion is not parallel everywhere throughout 
the area of the orifice, but is more and more oblique 
to that area as we pass from its center to its boundary. 
This converging motion of the elementary streams 
must make the pressure at the orifice somewhat greater 
in the interior of the jet than at its surface, and con- 
sequently the velocity in the interior of the jet must 
be less than that at its surface. 

Mathematical analysis is incompetent to determine 
either the motions or the actual paths of the issuing 
particles of liquid, but experiment shows that at a 
distance beyond the orifice equal to about half its 
diameter the converging motion ceases and all the 
particles have equal velocities in practically parallel 
lines. The reduced section of the jet at this point is 
called the ^Wena contracta,'^ and it is there that ex- 
periment shows the velocity to be very closely equal 
to that called for by theory, and given by the formula 

(61) The ratio of the cross sectional area of the jet 
at the vena contracta to that of the orifice is called 
the ^^ coefficient of contraction," and, as will be seen 
in the next chapter, varies with the shape and dimen- 
sions of the orifice and with the head of the liquid, 
having an average value of about 0,62 for orifices with 
sharp inner corners. That it cannot be less than 0.50, 
in the absence of friction, may be shown as follows: 



86 Hydbomechanics 

When the orifice is closed the pressure on it is 
(Po -|- wh)a^ where a is its area, h the depth of its center 
of gravity, po the atmospheric pressure, and w the 
specific weight of the liquid. When the orifice is 
opened, the only pressure opposing the exit of the 
liquid is p^a, and so there is an unhalanced pressure 
wha which acts to expel the liquid. Now, if c^ is the 
coeflQcient of contraction, the mass of liquid which 

w 

passes the vena contracta in imit time is — c^av, and its 

w w 

momentum is — c^av^ = — c^a X ^gh = 2wCiah; 

whence, by the fundamental relation Ft = mv, we 
have wha = 2wCiah, or c^ = 0.50. But really the 
opening of the orifice produces an unbalanced pressure 
greater than wha, since the motion of the fluid reduces 
the pressure on the walls of the vessel near the orifice. 
Consequently c^ must really always be greater than 
0.50. 

Experiment has shown that where a short cylindrical 
tube, projecting inwards, is attached to the orifice, 
thus causing the pressure on all parts of the walls of 
the vessel excepting the orifice itself to remain prac- 
tically the same with the orifice open as with it closed, 
the coefficient of contraction actually has its theo- 
retical value 0.50. 

Peoblems VI. 

(1) If the liquid which just fills a hemispherical 
bowl be made to rotate uniformly about the vertical 
radius of the bowl, how much will overflow? 

(2) A hollow paraboloid of revolution, axis vertical 
and vertex down, is half filled with liquid. What must 



Ai^ Elementary Treatise 87 

its angular velocity about its axis be in order that the 
liquid may just rise to the rim ? 

(3) A hemispherical bowl of radius a containing a 
given volume of water is set rotating about a vertical 
diameter. At what angular velocity does the water 
begin to overflow ? 

(4) If, in the preceding example, the angular velocity 

just exceeds \ — how much of the bowl is dry? 

(5) A narrow horizontal tube AB, with two vertical 
branches AC and BD, is filled with water to a given 
height. If this continuous tube be set rotating about 
a vertical axis through a point in AB, what will be 
the difference of level of the water in the two branches? 

(6) A closed cylinder just full of liquid rotates uni- 
formly about its vertical axis. What are the total 
pressures on its top, bottom, and on its curved surface ? 

(7) If a hollow open cone, axis vertical and vertex 
down, contain a given volume of liquid, discuss the 
question of whether it can be emptied by rotation 
about a vertical axis. 

(8) If the measured pressure and velocity at a given 
point in a pipe full of flowing water are respectively 
10 pounds per square inch and 18 f. s.; what is the 
pressure head; what is the velocity head; what, sup- 
posing there is no friction, would be the hydrostatic 
head if the flow were stopped ? 

(9) At points A and B in an outflow pipe the hydro- 
static heads are respectively 6' and 5', and the cross 
sectional areas are IJ and 2J square feet. If the pipe 
is discharging 16 cubic feet of water a second, what 
are the velocities, the velocity heads and the pressure 
heads at the two points? What is the pressure per 



88 Htdkomechanics 

square incli at A and at B before the flow begins and 
after it is established ? 

(10) A vertical vessel of hour-glass shape, full of 
water, has a cross section of .4 square feet at its 
smallest part, which is 4 feet below the surface, and 
is discharging 8 cubic feet per second from an orifice 
near its bottom. What is the pressure in the water at 
the smallest section, and what would happen if a small 
tube connected that part with another vessel of water 
at a lower level ? 

(11) If the hydrostatic pressure in a pipe were 80 
lbs. per square inch, what velocity must it have to 
reduce the pressure to 50 pounds per square inch? 
Water is flowing with a velocity of 25 f . s. in a pipe, 
how much will the pressure per square inch be 
increased if the flow is stopped ? 

(12) What is the velocity of efflux from a small 
orifice under a head of 6"; of 5'; of 100'? 

(13) What would be the velocity of efflux from a 
small orifice into a vacuum under a head of 6''; of 5'; 
of 100' ? 

(14) What would be the velocity of efflux from a 
small orifice under a surface pressure of 100 pounds 
per square inch and a head of 6"; of 5'; of 100'? 

(15) Water is being discharged through a small 
underwater orifice under a head of 60' on one side and 
10' on the other. What is the velocity of efflux ? 

(16) What is the velocity with which water under a 
head of 250' flows into a boiler in which the pressure is 
40 lbs. per square inch by gauge? 

(17) If the pressure in a boiler is 100 lbs. per square 
inch above the atmosphere, how fast will water fiow 
through a small orifice 2' below the water level ? 



An Elementary Treatise 89 

(18) How fast will mercury flow out of a small 
orifice under a head of 3' and a surface pressure of 
10 atmospheres? 

(19) A cylindrical vessel 3' in diameter and 4' deep, 
three-fourths full of water, is set rotating about its 
vertical axis until the water is just about to overflow. 
What will be the velocity of efflux from a small orifice 
in the base at a distance of 15'' from the axis? 

(20) A circular tube of radius a., half full of fluid, is 
made to revolve uniformly about a vertical tangent. 
Show that the diameter through the open surfaces of 

the fluid is inclined at tan"^ to the horizon. 

(21) A circular tube of radius a, containing a flla- 
ment of mercury which subtends an angle 2 a at the 
center, rotates about a vertical diameter. Show that 
the filament will divide at its lowest point when the 

angular velocity reaches a/- sec -«-• 

(22) An open vertical cylinder, height h, radius a, 

full of liquid, rotates uniformly about its axis, the 

peripheral velocity being v. Show that the vertex of 

v" 
the free surface is ^^ below the rim: that a volume 

—T — of liquid is spilt, and that when the motion is 

V- 

V" 

(23) Show that if, in the preceding example, —> h, 

the bottom of the cylinder will be uncovered in a 

/ 2^h\ 
circle of area-al 1 '—), and that when the cylinder 

is stopped the liquid will stand at a depth — %-. 



90 Htdkomechanics 

(24) If the cylinder is closed by a piston of weight 

P, and the weight of the liquid is W, prove that the 

v"* 2hP 

piston will rise when 7;— exceeds -^s^-. 
^ 2g W 

(25) At what angular velocity does the pressure at 
the lowest point of a closed sphere, full of liquid, and 
rotating about a vertical diameter, cease to be the 
greatest pressure on the surface ? 

(26) If a closed vessel full of fluid is rotating uni- 
formly about an axis, prove that the pressure which 
results from the rotation upon any portion A of the 

enclosing surface is -^ Ak^, where w is the angular 

velocity, p the density of the fluid, and Ak^ the mo- 
ment of inertia of the surface area A about the axis 
of rotation. 

(27) A hollow cone, vertex up, just full of liquid 
revolves uniformly about a vertical element. Prove 

that the pressure on the base is —^\ — (1 + 5 cos^a) 



tan a -|- 8 cos a , where W is the weight of the liquid, 
2a the vertical angle, and a the radius of the base. 



An" Elementary Treatise 91 



CHAPTEE VIL 
Discharge throug'li Small and Large Orifices. 

(62) In order to compute the actual discharge of 
liquid through an orifice we must know both the 
velocity and the cross section of the jet at some point. 
We have seen that the theoretical velocity of efflux is 
given by the formula v = \^2gh, and that under cer- 
tain conditions this velocity is very nearly attained, 
not at the orifice itself, bu.t in a contracted section 
of the jet just beyond it. 

It has been found that the form of the orifice 
materially influences both the amount of the contrac- 
tion of the jet and its velocity, and we will now 
consider what modifications of the formula for theo- 
retic discharge experiment shows to be necessary. 

(63) As already stated, the " coefficient of contrac- 
tion,^^ which will be called c^ , is the number by which 
the area of an orifice must be multiplied in order to 
find the area of the least cross section of the jet. 
Similarly, the "coefficient of velocity," designated by 
C2 , is the number by which the theoretical velocity 
of efflux must be multiplied to find the actual velocity 
at the least cross section of the jet. Lastly, the 
" coefficient of discharge," called c, is the number by 
which the theoretic discharge must be multiplied to 
find the actual discharge. 

^ Thus, if a is the area of an orifice discharging under 
a head h, and a' the area of the contracted vein; and 
if V and q are the theoretical, and v' and Q the actual 



92 



Hydkomechanics 



velocity of efflux and quantity discharged in unit time ; 
we have a' = c^a; v' = CgV = c^y^g^', c = c^Cg; and 
also q = av = a \/2gh ; Q = c^Cgav = ca\/2gh. 

The determination of the value of the coefficient of 
discharge (c) is alone necessary for the practical pur- 
pose of measuring the flow of a liquid, but a separation 
of this coefficient into its two factors c^ and c^ is 
necessary for the discussion of questions regarding the 
energy of jets. 

(64) An orifice so formed that the escaping liquid 
only touches its inner edge is termed a " standard 
orifice," and, on account of the regularity of the results 
given by such orifices, they are used, whenever prac- 
ticable, for the measure- 
ment of water. The object 
to be attained by a " stand- 
ard orifice " is to reduce 
the surface of contact of 
the jet with the vessel as 
nearly as possible to a line, 
and this result is attained, 
as shown in Fig. 18, either 
by making the orifice in a 
thin plate or by leaving the 
inner edge a sharp corner 
and beveling the outer edge of the orifice. 

With circular orifices of this character the vena 
contracta, or section of least area of the jet, occurs at 
a distance from the plane of the orifice about equal 
to its radius. At this point a jet under steady flow 
looks like a clear crystal bar, while beyond it the jet 
gradually enlarges and its surface becomes more and 
more disturbed. 




Fig. 18. 



An Elementary Teeatise 93 

With standard orifices of rectangular shape a similar 
contraction takes place^ the edges of the jet being 
angular and its cross section similar to the orifice 
until the vena contracta is passed. 

Experiments with water flowing through standard 
orifices show that the coefficient of discharge decreases 
slightly with increase of head and also with increase 
of size of orifice^ this being due to more perfect con- 
traction of the jet, since the coefficient of velocity 
increases with the head, probably often exceeding 0.99, 
Furthermore, the coefficient of discharge is found to 
be slightly greater for squares than for circles of the 
same diameter, probably due to imperfect contraction 
at the corners; to be greater for rectangles than for 
squares of the same area; and to be very slightly 
less for submerged orifices than for orifices discharging 
into the air. 

The mean value of the coefficient of contraction (c^) 
is about 0.62, and the mean value of the coefficient of 
velocity (c^) is about 0.98, making the mean value of 
the coefficient of discharge (c) about 0.61; and in all 
calculations of the flow of water through standard 
orifices these mean values can be used without any 
very material error. 

(65) If a rectangular orifice has an edge at the 
bottom of the vessel, the liquid flowing through its 
lower portion moves in lines perpendicular to the 
plane of the orifice and so there will be no contraction 
on the lower side of the jet. So, too, if the orifice is 
in a corner of a rectangular vessel, the contraction 
_will be suppressed on two sides of the jet, and, indeed, 
experiment shows that there is more or less suppres- 
sion of contraction whenever an orifice of any shape 



94 



Hydromechanics 



is close to the sides or bottom of the vessel. In such 
cases there is a somewhat increased discharge, but its 
amount varies greatly with the circumstances and is 
not well known, so that suppression of the contraction 
should be avoided where accurate results are sought. 

If the inner edge of the orifice be rounded, the con- 
traction of the jet is reduced, and the discharge in- 
creased; and, if the curve follows as nearly as possible 
the natural lines of contraction, as shown in Fig. 19, 
the issuing jet can be made to completely fill the 
orifice. This, however, is not really a 
suppression of the contraction, since 
the cross section at ab must bear about 
the same relation to that at cd as does 
the contracted vein to a standard ori- 
fice. With a carefully shaped mouth- 
piece of polished metal the mean value 
of the coefficient of velocity (Cg) is 
about 0.97, and, since there is no contraction, c^ = 1.00; 
and the coefficient of discharge (c) is also 0.97. 

Generally speaking, the more the direction of the 
liquid flowing in from the sides differs from that of 




FlG.19. 




C,= so ,S7 .6z .js- 

FlG.^0 

traction in case A closely approaches the minimum 
the main stream, the greater the contraction of the 
vein. Thus in Fig. 20, while the coefficient of con- 



An Elementaey Teeatise 95 

possible value 0.50; in case B it would be about 0.57; 
in case C about 0.63; in case D about 0.75; and in case 
E nearly 1.00. 

(66) When water is allowed to discharge through a 
short tube, or ajutage, 2J or 3 times as long as it is 
wide, the stream is uncontracted and non-transparent, 
and, while the velocity of efflux is less, the discharge 
is considerably greater than in the case of a standard 
orifice of the same area. With a shorter tube, the 
vein does not touch the sides of the tube, and so it has 
no influence on the efflux. With a longer tube, the 
stream may not fill the tube at first, but if its outer end 
is closed for an instant so that the tube is once filled, 
it will continue to run full, giving what is called 
" discharge of a filled tube." 

A " standard tube '' is one of sufficient length to 
have the escaping fluid Just fill its outer end, while its 
inner end is flush with the inner wall of the vessel 
and has a sharp edge like that of a standard orifice. 
With such a tube, since the Jet fills the outer end, the 
coefficient of contraction (c^) is 1.00, and so the coeffi- 
cient of velocity equals that of discharge. 

Experiment gives a mean value Cg ^ c = 0.83 for 
the standard tube, showing that its discharge is ff, or 
about 1.34 times as great as that of a standard orifice 
of the same cross section. 

The use of a standard tube 
made of glass shows that the Jet 
contracts in entering the tube Just 
as it does in issuing into the air 
from a standard orifice, and, if a 
small hole be made in the tube j^ _ . 

abreast the contracted vein a', in- T l u-. >6 1. 




96 Hydromechanics 

stead of water running out, air is sucked in. Further- 
more, discharge with filled tube ceases if air is freely 
admitted to the tube at a' (Fig. 21), and cannot be 
maintained under any circumstances when the head 
exceeds about 40 feet. 

(67) The practical interest in the circumstances of 
discharge with short tubes lies in the light which it 
throws upon the subject of the entry of water from 
reservoirs into conducting pipes, but it has also a 
theoretical interest as exhibiting what at first sight 
appears to be a contradiction of the law of the con- 
servation of energy. As stated, the coefficient of veloc- 
ity at a' (Fig. 21) is 0.82, and taking the coefficient of 
contraction at a' to be .62, this shows that the velocity 
in the contracted vein isff V2gh = 1.32V2gh, so that 
the energy of the jet at the contracted vein is {L32yh. 
= 1.74 times the theoretical energy due to the head 
h. The explanation, however, is simple, and lies in 
the fact that the flow of water causes a partial vacuum 
around the contracted vein, so that the atmospheric 
pressure on the surface of the water in the reservoir 
acts to increase the flow. Thus, with the figures 
already given, we find that 1.74 times the real hydro- 
static head is required to cause the actual flow in the 
contracted vein, which shows that there is an excess 
of pressure on the surface of the water in the reservoir 
over the pressure in the contracted vein equivalent 
to about three-fourths the hydrostatic head. Of 
course this excess of pressure could never in any case 
exceed the atmospheric pressure equivalent to 34 feet 
head, but even this is never attained since, when the 
hydrostatic head reaches about 40 feet, the reduction 
in velocity necessitated by the sudden expansion from 



An Elementary Teeatise 97 

the contracted vein to a full mouth, at atmos- 
pheric pressure could only occur with a negative pres- 
sure in the contracted vein, which is impossible. Con- 
sequently with a head of 40 feet or more a standard 
tube will not run full at the mouth. 

Eeturning, then, to the question of energy, we see 
that while the energy of the Jet at the contracted vein 
may exceed that due to the hydrostatic head, the 
excess must be expended in overcoming the atmos- 
pheric pressure in the outer part of the tube, so that 
in no case does the available exceed the theoretical 
energy. In fact the energy of a given quantity of 
water issuing in a jet depends solely upon its velocity 
of efflux, and since this velocity is 0.98 for a standard 
orifice and 0.82 for a standard tube, the energies, for 

Q QQ2 

the same weight of water, are as=^=^or as .96 to .67 

in favor of the orifice — 4:fc of the total energy due to 
the head is lost with the orifice as against 33^ lost 
with the tube. Even considering the energy of the 
jet per unit of time, regardless of the quantity of 
water used, we find a slight advantage on the side of 
the orifice, the two energies being about as .585 to .550. 

We would naturally expect the greatest efficiency 
where the resistance to efflux is least, that is with a 
standard orifice; and experiment confirms this view, 
showing that with large heads there is practically no 
loss of energy when a standard orifice is used. 

(68) The numerical values of coefficients thus far 
given are for water, and few experiments have been 
made to determine their values for other liquids. It 
may be said, however, that the more viscous a liquid 
the greater its coefficient of discharge. Thus a thick 



^8 



Hydeomechanics 



oil flowing through a circular standard orifice under a 
moderate head has a coefficient of discharge of about 
0.72; while mercury under the same conditions has a 
coefficient of discharge about 5^ less than that of 
water. 

(69) Hitherto we have considered only what are 
called small orifices — that is, orifices whose area is so 
small in comparison with the head that they can be 
assumed, without material error, to be under a uniform 
head equal to that on their centers. When an orifice is 
so large that there is considerable difference between 
the heads on its top and bottom, it becomes necessary 
to take account of the corresponding variations in the 
velocity of efflux. This is done by dividing the area 
into elementary strips and summing up the flows which 
would occur through them if they were separate small 
orifices. Of course this cannot give exact results, but 
it gives the closest approximation to the truth which 
can be mathematically determined. 

(70) As nius- 
trations of the 
method of de- 
ducing formulae 
for the dis- 
charge through 
orifices of large 
area we will take 
the two cases of 
rectangular and 
circular open- 
ings. 
Eeferring to Fig. 23, if we consider the rectangle as 
made up of elementary strips, each dy high and at the 




FTihi2 



An Elementaet Teeatise 99 

depth y, tlieii the discharge through one of those 
strips, regarded as a small orifice, is dQ = chdy X 
'\/2gj, and so the discharge through the whole rec- 

cbV^gy cly, the integration of which 
hi 
gives 

(18) Q = |cbV2^(h/-h,^) 
in which c is the coefficient of discharge and may be 
taken to be 0.61 when there is full contraction. 

Examination will show that this formula gives 
results slightly smaller than those given by the approxi- 
mate formula Q = cbdV^gh, where h is the head on 
the center of the rectangle, but that the difference 
is only 0.3^ when h = 2d and only about 0.1<^ when 
h = 3d. Accordingly the exact formula (18) need 
only be used when the head on the center of the orifice 
is less than twice its vertical width. 

Similarly, in the case of the circle, we have for the 
flow through an elementary strip, as shown in Fig. 22, 
dQ = 2cxdyVSg(h — y) = 2cV2gh Vr" — f 

y 1 - I dy. Whence Q = 2c v/2^ jV - f )^(l -~f 
dy, or expanding ( 1 — ^ j' by the binomial theorem, 



dy, each term of which is immediately integrable, 
giving 

(19) Q = c-rV"^^fl ^" ^^ 



32\f 1024h^ 
in which again c is the coefficient of discharge, having 
a mean value 0.61^ 
LofC. 



100 Hydkomechanics 

Here again the approximate formula, Q = C7:v^^/2gh, 
gives too great a result, but the error is only 0A% 
when h = 3r, and so the exact formula need only be 
used when the head on the center of the circle is less 
than three times its radius. 

(71) The principal practical application of formulae 
for discharge through large orifices is to weirs, or 
notches in the top of the side of a reservoir, through 
which water flows. Only rectangular weirs will here 
be considered. Usually the vertical edges of the 
notch are so far from the sides of the reservoir or 
feeding canal that the sides of the issuing stream are 
fully contracted, and such a form is called a weir with 
end contractions, while the form in which the vertical 
edges coincide with the sides of the feeding canal is 
called a weir without end contractions. 

- As shown in Fig. 23, the 

^^"-^■^l Mzi^ T^r-^^^ surface of the water takes 
Z3vlrvl/ ^_{Sr7^.^S5\ a curved form in approach- 
"^j-Trnlh 4 \Ut^ .^ f \ > ing the weir, so that the 
j^:^-'\ \^-^'i true head H must be meas- 

'?///7^^/A W<r^^^M ured some feet back from 

^ ^ ^ the crest, as the lower edge 

r I O//^, of the notch is called. For 

accurate results in the determination of the discharge 
it is also necessary that the crest and sides of the weir 
shall have definite sharp corners giving complete 
contraction. 

Taking now our formula (18) for a large rectangular 
orifice, and putting h^ = o, h2 = H, we have 

(20) Q = ScbV^Hi 

In which the value of c, found by experiments, varies 



An Elementary Treatise 101 

from 0.65 to 0.58, being greater the less the head, and 
in the case of weirs with end contractions increasing 
with the value of b, which is called the length of 
the weir. 

A mean value 0.61 may be nsed for contracted weirs 
and 0.63 for weirs with end contractions suppressed, 
but for accurate results the true value to suit the 
actual conditions must be obtained from a table of 
coefficients to be found in works on hydraulics. 

The value of H may vary from 0.1 to 1.5 feet in 
practice, and the length of the weir from 0.5 to 20 feet 
and upwards. 

(72) When there is an appreciable velocity of the 

water in the feeding canal at the point where H is 

measured, it must be taken account of, and this is 

done by regarding it as due to a fall h from the true 

level surface further upstream. The value of h is then 

v^ 
taken to be -^, where v^ is the mean velocity at the 

point where H is measured, this being called the 
" velocity of approach.^^ 

The best method of using this velocity of approach 
head in the formula for discharge through weirs is 
disputed. The most extensively used formula, that of 
Francis, takes h as head above the upper edge of the 
weir, H + h being the head on its crest, and, for weirs 
without end contractions, is 

(21) Q = 3.33b [(H + h)^ — h^] 

in which 3.33 is a mean experimental value of fc V2g, 
and which is modified, for weirs with end contractions, 
by using b — 0.2 H instead of b for the length of the 
weir. 



102 Hydeomechanics 

Smith's formula is 

(22) Q = fcbV2^(H + nh)? 
the value of c being given in his tables, and, as already 
stated, varying from 0.58 to 0.65, according to cir- 
cumstances, and n being 1.4 for weirs with full con- 
traction and 1.33 for weirs with end contractions 
suppressed. 

Formula (21) is the more correct in form, while 
formula (22) takes account of the fact that the central 
current of the stream has a velocity greater than v^, 
which is the mean velocity of approach. In good 
practice the velocity of approach should be as small as 
possible, not exceeding 1 f. s., and it is only occasion- 
ally, and in the case of weirs, that it needs to be* 
considered. 

(73) In the case of orifices, fortunately, the velocity 
of approach is almost invariably so small as to exert 
no appreciable effect on the discharge. Should it be 
desirable to allow for it, a consideration of the methods 
of deducing the formula v^ = 2gh, both of which 
assume a zero velocity at the surface, indicates that 
where that velocity is not zero, but has a value v^ , we 
must put v^ = 2gh + ^^1 • S^t if ^ l>6 the cross 
section of the vessel at the surface, and a the orifice 

area, v. = -r- . Whence we have — 
A 



. 2gh 
(23) 



^-(1)' 



Thus, as was to be expected, the discharge is least 
when there is no velocity of approach, and, as will be 
found upon trial, when the area of the orifice is -^ 



An Elementaey Teeatise 103 

the cross section of the vessel^ the effect of the velocity 
of approach ceases to be appreciable. 

(74) We are now in a position to calculate the 
quantity of water which will flow from any given 
orifice in a given time. It is only necessary to be 
careful in applying the formulae to use the same units 
for all the factors. Thus if g is given the value 32, 
which is feet per second, h must be put in feet, and a 
in square feet, when Q will be found in cubic feet 
per second. 

As an example, let us calculate how long it will 
take for a rectangular pontoon, 20' X 10' X ^'^ which 
draws one foot when empty, to be sunk by the entry 
of water through a hole of 18 square inches area in its 
bottom. Here there is a constant difference of head 
of 3' between- the water without and that within the 
pontoon as the latter gradually settles, and, when 
20 X 10 X 3 ^ 600 cubic feet of water have entered, 
the reserve buoyancy is wholly exhausted, and the 
pontoon will sink. The influx per second is Q = 
.61 X t'A V^ X 32 X 3 = 1.055 cubic feet, and so 

to sink. 



it will take ^ = 568.8 seconds = 9 m. 48.8 sec. 



Peoblems VII. 

(1) How much water will flow through a standard 
opening 8 square inches in area, under a head of 20 
feet, in 5 minutes ? 

(2) If 325 cubic inches of water per second flow 
through a standard opening of 5 square inches, what is 
the head ? 

(3) How many cubic feet of water will flow through 



104 Hydromechanics 

a standard orifice of one square inch in an hour, if the 
water level is maintained 9 feet above the orifice? 

(4) Deduce a formula for the discharge through a 
triangular orifice, altitude a and base b, when the ver- 
tex is in the surface and the base horizontal. 

(5) Deduce a formula for the discharge through a 
triangular orifice, altitude a and base b, when the 
base is in the surface. 

(6) Deduce a formula for the discharge through a 
segment of a parabola, height a, base 2b, when the base 
is in the surface. 

(7) Deduce a formula for the discharge through a 
segment of a parabola, height a, base 2b, when the 
vertex is in the surface and base horizontal. 

(8) Calculate the discharge per minute through a 
rectangular orifice, 2 feet wide and 1 foot high, with 
its upper edge IJ feet below the surface. 

(9) Calculate the discharge per minute through a 
circular orifice of 6" diameter with its center 16" 
below the surface. 

(10) A rectangular pontoon 18' X 9' X 3', drawing 
6" when empty, has a hole of 9 square inches area in 
the bottom. How long will it take to sink? 

(11) What will be the discharge per second through 
a circular orifice a foot in diameter with its upper 
edge at the surface? 

(12) What will be the discharge through a tri- 
angular orifice of 18'' base and 6" height, the base 
being in the surface ? 

(13) What will be the discharge through a tri- 
angular orifice of 6" base and 18" height, the vertex 
being up and at the surface and the base horizontal? 

(14) A small stream is dammed with boards, through 



An Elementaky Treatise 105 

which a rectangular hole V wide and 6" deep is cut. 
The water ceases to rise further when it stands 1' 
above the top of the orifice. What is the flow of 
water in the stream ? 

(15) What is the discharge per second over a weir 
6' long, the head being 0.465 feet and the velocity of 
approach being negligible ? 

(16) Compare the discharges calculated by Francises 
formula and by Smithes for a weir, with end contrac- 
tions, 4' long, the head being 0.427 feet and the 
velocity of approach 1 f. s., using c = 0.61. 

(17) The floating dock at Algiers will support an 
18,000-ton ship with the floor of the dock just awash, 
and has 3' freeboard when carrying a 15,000-ton ship. 
How long does it take to sink it from the second to 
the flrst position by the influx of water through 16 
circular valve openings, each of 16'' diameter, in the 
bottom of the dock, which is 17' 6" below its floor? 

(18) The quantity of water discharged through a 
standard orifice of one square inch area under a head 
of 6 ''5 is known as a " miner^s inch." If the discharge 
through an orifice 250" long by 4" high, under a head 
of 9" on its center, is sold for 1000 miner^s inches, 
what error is made ? 

(19) What error is made in selling the discharge 
through a rectangular orifice 12" deep and 12'' 75 wide, 
with a head of 6" above the upper edge, as 200 miner's 
inches ? 

(20) A hole of one square inch area is made in a 
boiler under a steam pressure of 180 pounds per square 
inch. What will be the approximate discharge of 
water per second? 

(21) Water is discharging through a standard tube 



106 Hydromechanics 

under a head of 40'. What are the velocity and pres- 
sure heads at the mouth of the tube; what are they at 
the contracted vein; how do you account for the loss 
of head at each of these points ? 

(22) A vessel of 60 square inches cross section is 
kept full of water which is discharging through a cir- 
cular orifice 5" in diameter under a head of 8'. What 
is the velocity at the surface and what the discharge 
per second? 

(23) In order to measure the flow of water in a 
canal 5' wide, a transverse partition 2' high, with 
sharp upper edge, is placed in it, and when steady flow 
is established, the height of the surface is found to be 
1/5 above the sill. How much water per second flows 
through the canal? (Use Smith's formula with c = 
0.64 and n = 1.33.) 

(24) An empty, air-tight, rectangular box, 8' high, 
is placed with its top 12' below the surface of the sea. 
If a small hole is made in its bottom, how high will the 
water rise in the box? 



An Elementaky Treatise 



107 



CHAPTEE VIII. 

Time of Emptying Vessels— Efflux of Gases — ^Flow 
Through Pipes. 

(75) When a vessel, or 
reservoir, receives no in- 
flow of water while it is 
being emptied through an 
orifice, the head is a drop- 
ping one, and not con- 
stant. 

Let Fig. 24 represent a 
vertical section, through a 
small orifice at o, of a 
vessel of any shape. Take 
axes as shown; suppose the orifice area to be a; and 
let K = f(y) be the surface area at the distance y 
above the origin. Then, if dy be the drop of the 
surface in the time dt, we must have — Kdy = 
caV^gy dt, each of these expressions being the dis- 
charge in the time dt (dy is negative because y de- 

.-f(y)dy 




Fig- £4. 



creases as t increases). Hence we have dt = 



Jo Jhca/v/^gy J" 



h ca v/2gy 
(18) r = P 



f(y)dy_ 

o ca \/2gy ' 
■^ %)dy 



ca V'2gy 



, ca A^2gy 

In which r is the time corresponding to the fall of 
the surface from the height h to the orifice, or the 
time of emptying the vessel. 

Thus, for example, take a cylindrical vessel, of height 



108 Hydromechanics 

h and constant cross section A, being emptied through 
an orifice of area a in the bottom. Then we have r = 

— — -== = a/ ?r- • This is twice the time it 

caJox/2gy ca M 2g 

would take to discharge the same quantity under a 

constant head h, for in that case the discharge per 

second would be caV^gh, and, the contents of the 

cylinder being Ah, the time to discharge its contents 

Ah A /T" 



= 4\/- 



would be -, ^ 

ca \/2gh ca M 2g 

As another example, take a sphere of radius r, with 

a small orifice of area a at its lowest point. Here, 

again taking the origin at the orifice, we have for the 

variable surface area K = ttx^ =. 7r(2ry — y^). Hoiice 

ca-v/^gJo \/y caV^g 13 5 Jo 

167rri 



15ca x/g 

(76) If, on the other hand, it were desired to de- 
termine the time during which the surface will fall 
from the height h^ to the height h^ , it is only neces- 
sary to integrate between those limits instead of be- 
tween the limits h and o. Thus, in the case of a 
cylinder of cross section A, we have, for the time from 
height h^ to height hg of water above the orifice, t = 

2A 
r= (\^ — \^). Moreover, from this equation, by 

transposing and squaring, we get Q = A(h2 — h^) = 

tea f \/2gh2 — -^-r- 1, where Q is the discharge in t 

seconds; whence the discharge in the first second is 

seentobeca(v/2p;-|g). 



An Elementary Treatise 



109 



(77) The emptying and filling of a canal lock affords 
a practical example of the use of the formula for 

discharge under a vari- 
able head. The lock B 
(Fig. 25) is filled by 
opening one or more 
orifices in the upper 
gate, which separates 
it from the "head 
bay" A, and is emp- 
tied through orifices 
in the lo"^er gate into the ^' tail bay " C. Conse- 
quently, both in filling and in emptying, the head on the 
orifices varies from H to o. Let the constant surface 
area of the lock be G-, and the orifice area be a. Then 

G f° dy 




Fig. 25. 



the time to fill the lock is ■ 



G 



to empty the lock is 



ca JH\/2gy 
o_dy^ 



and the tune 



integrals equals— y — 



ca J HV^gy' ' 



each of which 



2g' 



Thus if G = 80' X 30' = 1600 square feet; H = 

12'; and a = 8 square feet; we have r = — = 322 

seconds = 5 min. 22 sec, if c has its value (0.62) for 
sharp cornered orifices. But if the inner edges of the 
orifices are rounded and c is increased to 0.80, we have 
the time of filling or emptying the lock reduced to 
4 min. 10 sec. 

(78) It was once customary to measure intervals of 
time by a water clock, or clepsydra, which consisted 
merely of a vessel from which water escaped through 
a small orifice, the fall of the surface indicating the 



110 Hydeomechanics 

elapsed time. Assuming such a vessel to have the 
form of a surface of revolution about a vertical axis 
through the orifice, and taking x as the radius of a 
transverse section at the height y above the orifice, 

we have — = dt, or ^ oc -^ , whence we see 

ca x/ 2gy dt x' ' 

that if a vertical section of the vessel has the equation 

dy 
y = kx*, -7^ will be constant and so the surface will 

descend equal distances in equal times. 

(79) The laws of efflux thus far set forth apply 
only to liquids, their theoretical deductions having 
assumed that the density of the fluid considered was 
constant, and their empirical modifications being based 
upon experiments with liquids. We will therefore 
briefly consider what modifications of those laws are 
necessary in order that they may apply to gases. 

Eeturning then to the deduction of Bernoulli's 
theorem in Chapter VI; and supposing that a stream 
tube in a mass of gas, and not of liquid, in steady 
motion, is being considered, it will still be true that 
the same mass must enter the stream tube at P in 
a unit of time as leaves it as P^ in the same interval, — 
we still have /?vo- = Pi^i'^v only now p and p^ differ 
whereas in a liquid they are equal. This change of 
density of ttie gas due to its passage from P to P^, 
however, introduces an essentially new element into 
the problem, since the gas, by virtue of its expansive 
power, possesses a so-called intrinsic energy which 
is a function of its density, and which therefore is 
greater or less at P^ than at P depending upon whether 
it is more or less compressed there. Now the intrinsic 

energy of unit mass of any gas is pdv where v is the 

Jo 



An Elementary Treatise ill 

volume of unit mass and this, of course, is the same 

thing as — Pf^f—J- Hence in our statement of the 

equality of the energies flowing into and out of our 
imaginary stream tuhe, we must add to one side the 
intrinsic energy of the entering gas and to the other 
side that of the departing gas. Thus we have 

x+^ + ,_ifV( 



gi^i 2g g Jo^ \ /> / 

Now, if we assume that the change of pressure from 
p to pi takes place without gain or loss of heat, or is 
an adiahatic change, we have p = ^' p^, k' heing a 
constant and y the ratio of the specific heats of the gas, 

whence — [ pd( — ) = k' p'^'^p = -^ = - — ^^— r- . 

J V ^ / J r - 1 (r — l)iO 

This reduces our equation to 
p v^ P 



%p ^g g(r ~ i> 



Pi ^1 



g/>i 2g " g(r-iK 
(19) -_^:l^h_z1+, 

gp{:r - 1) 2g 

- g9i:r - 1) + 2g + ^^ - ^ • 

Thus we see that Bernoulli's theorem applies to gases 
subject to adiahatic changes with only the substitution 

of ^ ^ . -^ for the term -1-. 
r — 1 giO gp 

(80) Applying this theorem, now, to the case of the 

efflux of a gas from a large reservoir at pressure Pi 

into another where the pressure is ps , through a small 



112 Hydeomechanics 



orifice, we can consider v^ as zero, and so get for the 



yelocity of efflux, -^ = — ^ ( JlL,- J^ ] + z, — z, . 

Bnt (z^ — Z2), the difference of the heads at a point 
inside the vessel and at the orifice, is so small in com- 
parison with the other part of the expression that it 
may be neglected. Moreover, since p = kV^, we have 

~ = I— ]y , or p^ = pJ-^ ]y. Hence we have, finally, 

2g gMr- i)L \vJ J ^r — 1 

and calling the velocity of efflux v, 

(.0) v = ^-^[i-(A)y 

in which w^ is the weight of unit volume of the gas at 
pressure p^ and n = ^^ = 3.541, and which reduces 

for air to v = 2413 VI + «t \/ 1 - 1-^)' "f. s., or, for 

t = 20° C, to 2510 y 1 — (^y\ s. 

(81) Of course (20) should reduce to the expression 

v2 = 2gf-^-^ ^j, corresponding to v^ = 2gh for 

liquids, when p^ and p^ are so nearly the same that 
the density of the gas within and without the vessel 
may be considered equal, and, to show that this is so, 
let P2 = p^ — X, where x is small compared with p^ . 

Then (20) becomes v^ = ^^^ fl - (l -— )^1. and 
since f 1 p= 1 '+..., we have, neglecting 

X 

higher powers of the fraction — , 

Pi 



An Elementary Treatise 113 

This last formula may be used without great error 
whenever the difference of pressures is not more than 
one-fifth that of the reservoir. Thus if p^ = 18 and 
P2 = 15 pounds per square inch, and if the tempera- 
ture be 25° C, then w-^, the weight of a cubic foot of 
air in the reservoir, is if X f|f X .0807 = .0887 

A WT. T. 2 64 X 144 (18 - 15), 

pounds. Whence we have v^ = ^r^^^ -^' 

.Uoo7 
96 
and V = .— — - = 558 f. s. While the more accu- 
V.0296 

rate formula (20) gives, with the same data, a result 

about 3-J^ greater. 

(82) The coefiicients of velocity and contraction for 
gases have not been well determined, and such experi- 
ments as have been made seem to indicate that they 
vary more widely than in the case of liquids. Weis- 
bach recommends as coefiicients of efilux for air, 0.56 
for standard orifices; 0.75 for standard tubes; and 0.98 
for well-rounded mouthpieces, and probably the coeffi- 
cients determined for water can be used for air in 
all cases without very great error. 

(83) We have assumed in the deduction of our 
formula for the velocity of efflux of gases that the 
pressure in the contracted vein is the same as that 
of the outside medium, and that the issuing jet has 
reached this reduced pressure adiabatically. Hence, 
to get the weight of gas discharged in unit time, we 
must multiply the volume discharged (cav, where c is 
the coefficient of efflux and a the area of the orifice) 
by W2 , the weight of unit volume at the outside pres- 
sure. But W2 = Wj ( — jY when the change from p^ to 



a maximum. 



114 Hydromechanics 

P2 is adiabatic, and so, calling — = x, we have for 

Pi 
the weight of gas discharged in unit time — 

(21) W == caw.xv y 2ng^(l - xA . 

Now, regarding this as a function of x, it will be 

dW 
found by putting — .j — = o, that the value x = 

(2-iiT-/= (rf -/= '■''' °^^'^" "^ 

In other words, whatever value the pressure in the 
reservoir has, the maximum possible weight of gas will 
be discharged per second when the pressure of the 
outside medium is a little over half that in the 
reservoir. 

It is remarkable that experiments seem to show that 
regardless of what the outside pressure may be, and 
even though the discharge be into a vacuum, the 
pressure in the contracted vein never falls below 
0.527 pi, so that, if the outside pressure is less than 
0.527 times the inside pressure, the velocity of efiiux 
assumes a constant value, found by putting x =: 0.527 
in (21), and which for air is 997V 1 + "t f- s., or at 
20° C, 1036 f. s. Under the same conditions the 
weight of gas discharged per second is kcaw^ pounds, 
where k is a constant having the value 520 for air. 

(84) One of the most important practical problems 
in hydromechanics is the determination of the relation 
which exists between the dimensions of a conduit or 
pipe and the quantity of water which it will deliver 
under a given head. 

We will first consider the simple case of a uniform 
pipe discharging with full section at its end. Let H 



Ax Elemextaey Teeatise 115 

be the head which causes the flow, bfeing the difference 
of level between the water surface in the reservoir 
and the center of the end of the pipe if the discharge 
is into the atmosphere, or the difference of level of 
the two water surfaces if the outlet is submerged. 
Let V be the velocity of discharge, which, since the 
cross section of the pipe is uniform, is also the velocity 
of flow throughout the pipe. Let d be the inside 
diameter of the pipe; and let 1 be its total length, 
following all its bendings. Then, if h^ is the head 
which ^vill produce the velocity v, and if h^ is the 
total loss of head due to resistance in the pipe, we 
must have H = h^ -{- ^2 • 

We have seen that under a head less than 40 feet a 
short tube placed just flush with the inner wall of a 
reservoir runs full at the mouth with a velocity v = 

cV^srh, whence we have h = ~ — ^ for the head which 
o 2gc'' 

will produce the velocity v in a pipe, the value of c 
being 0.83 for a sharp cornered entrance and approach- 
ing 1.00 for a well-rounded one. We shall assume c == 
0.82 for all cases, and therefore have 

(32) h. = 1.5^. 

The pipe being of uniform diameter, the resistance 
to flow through it can only be due to bends, obstruc- 
tions, and friction, and we will consider separately 
these three factors in the total loss of head in the pipe. 

The loss of head caused by an easy curve is inap- 
preciable, and even for a sharp curve is small, being 

v' 
stated by Weisbach to be onlv 0.3—- for a 90° chancre 

of direction of the axis of the pipe in a curve whose 



116 Hydromechanics 

radius equals the inner pipe diameter. So this part 

of hg need only be considered when there are a number 

of sharp bends, which should never happen in good 

practice. 

The loss of head caused by a partly closed valve may 

v' 
be expressed in the form k^, in which k, as given by 

Weisbach, is 5 or 6 for a valve about one-third closed 
and may reach several hundred if the valve is nearly 
shut. As our calculations are only to be applied to 
pipes running full flow, this cause of resistance will 
be considered non-existent, and it is only referred to 
as indicating the very serious loss of head which may 
result from an accidental obstruction in a pipe. 

By far the greatest resistance to the flow of water 
in a long pipe is caused by friction against the walls 
and accompanying eddies and cross currents, and it 
is found by experiment that the loss of head from 
this cause is directly proportional to the square of the 
velocity of flow and the length of the pipe, and in- 
versely proportional to the pipe's diameter. 

We can therefore write for the loss of head due to 

all resistances in a uniform pipe of easy curves, 

f 1 v' 
(23) K=j.^ 

in which f is a coefficient depending for its value 
primarily upon the roughness of the inner pipe sur- 
face, but which is also found to diminish with increase 
of the velocity of flow and with increase of the diameter 
of the pipe. 

v^/ fl\ 

Thus we have H = f 1.5 + J, from, which we get 



An Elementaey Tkeatise 117 



(24) v= /_ML^ 

and this formula should be used for comparatively 
short pipes, v being first determined approximately by 
the use of an estimated value of f, and then more 
accurately by using the tabular value of f correspond- 
ing to the velocity of flow first found. 

If the length of the pipe exceeds one thousand of 
its diameters, we may neglect the 1.5 in comparison 

with -V and write 
a 



(25) v^V-^ 

(85) With clean iron pipes, either smooth or coated 
with coal-tar varnish, f varies from .01 for large pipes 
and high velocities to .04 for small pipes and low 
velocities. It may be assumed to be .03 for diameters 
from 0''5 to 3V0, and .02 for diameters from l.'O to 2.'0, 
for clean pipes and velocities from 3 f. s. to 6 f . s. 
For accurate work, however, its appropriate value must 
be sought in tables to be found in works on hydraulics. 

With very foul pipes the values of f just given should 
be increased 50;^, or even 75^. 

(86) To determine the discharge through a long 

pipe we have Q= ^ v, or 

and from (26) we find for the diameter of pipe which 
will discharge a s^iven quantity of water (Q) 



118 Hydromechanics 

Also, for the head necessary to give the discharge Q 
through a given pipe, 

(38) n-^j^. 

Thus we see that if the diameter of a pipe is doubled 
the quantity of water it will deliver is increased in 
the ratio V^^ : 1, or nearly sixfold; while the head 
required to furnish a given quantity of water through 
a pipe of given diameter is 32 times as great as if that 
diameter were doubled. 

As a practical example, let us calculate the discharge 
of a pipe V in diameter and 10,000' long under a head 

of 100'. Taking f = .02, we have v =J^^2^^^9>^ 
^ ' > .02 X 10000 

= y'M = 5.65 f. s. and Q — ^v = 4.44 cubic feet 

per second. The head at the discharge end is there- 

fore only — = 0.5 feet, showing that 99.5 feet is the 

head used up in overcoming friction in the pipe. 
(87) The formula for velocity in a long pipe, when 

put in the form v = n y —. -r- and written 



(29) V = nVR. S. 
is known as Chezy's formula, and is the one most 
used for water conduits of all sorts. In the Chezy 
formula R is called the "hydraulic radius," or the 
" hydraulic mean depth," and is found by dividing the 
cross section of the stream by the wetted perimeter 

of the conduit, this being -^ = -^ for a full pipe of 



An Elementaey Tkeatise 119 

circular section; while s =^ is called the "hydraulic 

inclination/^ h being the total head diminished by 
the head required to produce the velocity in the pipe 

f H — - — ^ L and 1 being the total length of the pipe. 

The constant n is^ of course, experimentally deter- 
mined, and includes V^^^ with g expressed in feet per 
second, so that in the use of Chezy's formula E must 
be expressed in feet. 

For circular pipes, or semi-circular open conduits, 
having smooth inner surfaces, and for moderate veloc- 
ities, n may be given the values 98, 114, 126, and 140 
for diameters of 1'', V, %' and 4' respectively. 

As a practical example, we will find the discharge 
through a pipe %" in diameter and 96' long under 
a head of 5'. Here, as an approximation, v = 98 



I 



24 X 96 



4.56 f. s. ; from which we have for the 



v' 
head which will produce the velocity h-^ = ^ — ^ = 

'^l_'^_p:^ = 0.49 feet. Therefore H — h^ = 4.^51, and 
04 



^ = ^'^ V a/?96 = *-^* *• '•' '^^'''""' "^ = IS ^ 

4.34 = .095 cubic feet per second. 

With the same data formula (24) for short pipes 
gives V = 4.13 f. s. and Q == 0.090. 

Pkoblems VIII. 

(1) Find the time of emptying a cone with vertical 
axis through a small orifice in its vertex. 



120 Hydbomechanics 

(3) Find the time of emptying a cone with vertical 
axis through a small orij0.ce in its base. 

(3) Find the time of emptying a hemispherical bowl 
through a small orifice at its lowest point. 

(4) A closed hemisphere full of water stands with 
its flat side down; how long will it take to empty it 
through a small hole in the base? 

(5) A vessel formed by the revolution of a semi- 
cubical parabola (ax^ == y^) about its vertical axis is 
filled with water to the point where the radius of the 
surface equals its height from the vertex. How long 
will it take to empty through a small hole at the 
vertex ? 

(6) Water stands 20' deep in a vertical cylinder of 
50 square feet cross section. How long will it take for 
the surface to fall 10' by reason of the discharge 
through an orifice of 9.5 square inches in its bottom; 
and how long will it take to then fall 5' more ? 

(7) How long will it take to empty a sphere of 4' 
radius through a hole of 1 square inch at its lowest 
point ? 

(8) The ditch around a fort is a mile long, 30' wide, 
and 9' deep, and is full of water. How long will it take 
to drain it to a depth of 3" through a vertical cut ?J 
wide? How long will it take to lower the surface 
another inch? (Assume opening to have well-rounded 
sides so that there is no contraction). 

(9) A vessel is of the form given by the revolution 
of y = 64x* about the axis of Y, and has a circular 
orifice of 1" diameter at its lowest point. What ver- 
tical fall of the surface of water contained in it will 
measure an elapsed time of one second? 

(10) Two vessels of uniform horizontal cross sec- 



An Elementary Treatise 121 

tions^ respectively A and B, communicate by a small 
submerged orifice of area a. If the levels differ by h 
feet, how long will it take for them to come to a 
common height? 

(11) How long does it take to fill and how long to 
empty a single lock of mean dimensions 200' X 24', the 
filling orifice being 2/5 wide and 4' high and its center 
5' from both water surfaces when the lock is empty; 
and the emptying orifice being 2/5 wide and 5' high 
and entirely submerged? (Use c = .62). 

(12) A large reservoir contains air at 120° C. and 
17 i pounds pressure per square inch. What will be 
the velocity of efflux into the atmosphere and what 
the volume and weight discharged per second through 
an orifice of 2 square inches ? 

(13) What is the velocity and weight per second of 
discharge of air at 22.5 pounds per square inch and 
20° C. through an orifice of 1 square inch area into 
the atmosphere? 

(14) A Whitehead torpedo flash contains 11.77 cubic 
feet of air at 1500 pounds per square inch. How long 
vdll it take for discharge through a circular hole of 2" 
diameter to reduce the pressure to 50 pounds per 
square inch? 

(15) A reservoir contains natural gas at 30 pounds 
per square inch and 20° C, weight of the gas being 
.0484 pounds per cubic foot at 0° C. and atmospheric 
pressure. What quantity and what weight will be 
discharged per second through a short tube of 4" 
diameter into the atmosphere? 

(16) A uniform pipe of 3" diameter and 120' long 
discharges water under a total head of 24'. What is 
the velocity in the pipe ? 



122 Hydeomechanics 

(17) What should be the diameter of a pipe to dis- 
charge 1200 gallons a minute at a point a mile distant 
from and 96 feet below the surface of a reservoir? 
(1 gal. = 231 cu. in.) 

(18) What total head is needed for the discharge of 
500 gallons a minute through a 6" pipe a mile long? 

(19) A pipe 400 feet long and 8" in diameter dis- 
charges under a total head of 32 feet. What are the 
velocity and the discharge per minute if the pipe is 
straight and with no obstructions, and what are they 
if bends and a partly closed valve cause a loss of head 
of 4.0 times the velocity head? 

(20) If the pipe of the preceding example were 
4000 feet long, what would the velocity and discharge 
be under each of the two circumstances stated? 

(21) What is the discharge in gallons per minute of 
a 1" pipe 50' long under a head of 12' ? 

(22) What must the average inclination of a pipe be 
in order that the velocity of flow may be .82V2gh 
where h is the head on the entrance ? 

(23) What must be the diameter of a pipe 1000 feet 
long to discharge 3 gallons a minute under a head of 
11.3 feet? 

(24) A circular masonry conduit 4' in diameter runs 
half full and has a slope of 1.'5 in 1000'. Compute 
its discharge by Chezy's formula, using n = 120. 

(25) What must be the dimensions of a rectangular 
trough, of width double its depth, and with a slope of 
.002, in order that it may deliver 120 cubic feet of 
water a minute when running full, assuming n = 100 ? 

(26) What should be the diameter of a conduit to 
deliver 12,000,000 gallons of water a day, the slope 
being .00016 ? 



An Elementaey Treatise 123 

(27) A village street has 100 houses which draw 
water from a main 3000' long which is supplied from 
a reservoir 100' above the dead end of the main and a 
mile from its beginning. What must be the uniform 
diameter of supply pipe and main to be able to furnish 
water with a pressure head of 75' at twice the average 
rate of consumption^ which latter is 500 gallons a day 
for each house? 



124 Hydkomechanics 



CHAPTEE IX. 

Compound Pipes — Water Power — Energy and 
Reaction of Jets. 

(88) Thus far we have only considered "iiniform 
pipes,, and changes of section must sometimes he taken 
account of. 

If Vi and V2 are the velocities in adjacent pipe sec- 
tions of areas respectively a^ and ao , and if there were 
no loss of head in passing from the section a^ to the 
larger section ag , the pressure head in ag would be 
greater than that in a^ by the decrease of velocity 

head, or ^ — —. But actually the pressure in ag 

must exceed that in a^ just enough to cause a retarda- 
tion Vj — Vg of the mass of water /^a^Vi which enters 
it each second. Consequently the pressure intensity 

1 XT- J. • -u P^\^\ (^1 ~ ^2) 
m ag must exceed that m a^ by ■ — ^ = 

/>Vi(Vi — V2); and the head corresponding to this is 

/>v,(v, - vj ^ v/v, - vj ^ ^^^^ ^^^^^ ^^^ ^^^^ _^ 

gP g 

loss of head '^(^^i^Zl^ - ^^^^ = -^^V^^^ 
g 2g 2g 

which, since a^v^ = agVg, may be written in either of 

the following forms : — 

It is this loss by sudden expansion which explains 
the fact, already mentioned, that a jet issuing from a 



An Elementaey Tkeatise 125 

standard tube has so much less energy than one issuing 
from a standard orifice. In the second case the energy 
per unit weight of water is Cs'h = 0.96h, since the 
coefficient of velocity c^ = 0.98; while in the first 
case C2 = 0.82, and the energy is 0.67h, or 0.29h less. 
Now as there is a loss of 0.04 with the standard orifice, 
we may assume an equal loss due to the contact of the 
jet with the mouth of the tube, and this leaves 0.25h 
to be accounted for by the expansion of the jet from 
its contracted section (0.62a) to the full section of the 
tube (a). But formula (30) shows this to be correct, 

"62 ~ '^) 2'cr ^ ^-^^^ ^ -^^^ "" ^•^^^^• 

(89) A sudden decrease in the section of a pipe 
causes a similar, but smaller loss of head due to the 
expansion of the stream after it has been contracted 
in passing from the large into the small section. Thus, 
if c^ be the coefficient of contraction of the stream, 
aiid V2 its velocity after reexpansion to fill the pipe, 
the loss of head, by formula (30), is given by 

and, since the maximum value of c-^ would be 0.62, 

v' 
this loss cannot exceed 0.377t^. It should be noted, 

^ . ■ 
however, that y^ in this case is the velocity in the 

small section of the pipe, which may be much greater 

than Vi, so that the loss of head may be many times 

the velocity head in the large section of the pipe. 

(90) In good practice any necessary change of sec- 
tion in a pipe is made gradually, by means of a 
reducer, and does not cause any loss of head, but should 
there be, in any case, losses of head due to abrupt 



126 



Hydromechanics 



changes of section, numerous bends, partially closed 

valves, or any other cause than friction, they must be 

taken account of by summing them up in the expres- 

v' 
sion k —7 and then using the formula 



(32) 



f 



2gH 



5 + 



+ k 



(91) If, at intervals along a pipe running full flow, 
vertical tubes, open at each end, are placed with their 
lower ends flush with the upper inner surface of the 
pipe, the heights of the water surfaces in such pipes. 




which are called open " piezometers,^^ will evidently 
be the pressure heads at the corresponding points in 
the pipe, and a line joining these surfaces is called the 
" hydraulic grade line,'^ or the " hydraulic gradient.'^ 
When the pipe is laid in an approximately straight 
line, and the only resistance is that of friction, the 
hydraulic gradient is also a straight line, its height 
above the pipe diminishing in direct proportion with 
the length of the pipe from its entrance. Thus, in 
Fig. 26, CE is the hydraulic gradient, beginning h^ 
below the surface A and ending at the surface E. The 



An Elementary Treatise 127 

distance BE is the total head H, of which DE = h2 
is lost in overcoming resistances in the pipe^ and BD 
= AC = h.^ is the head required to produce the 

velocity of flow v, or -r — -, . At the entrance the 

-^ ' 2gc' 

pressure head is OC, and at the outlet 0' it is O'E; 

v' 
while the velocity head throughout is —, which is less 

than h^ by the loss of head at entrance. 

(92) It is necessary for continued full flow of a 
pipe that no part of it shall rise above the hydraulic 
grade line. Thus, if the entrance were above the 
point C, the head upon it would be less than h^ , and 
so would be insufficient to fill the pipe; and, if at any 
other point the pipe rises above CE, the pressure in 
it at that point will be less than the atmospheric pres- 
sure; air will collect there, and a great diminution of 
the discharge will result. 

When a large part of a pipe conveying a liquid lies 
above the hydraulic gradient it is called a " siphon," 
and for the continued successful action of such a 
device it is necessary to furnish means, such as an air 
pump, for removing the air which will gradually 
accumulate at the point of greatest elevation. 

(93) In the year 1890 about 6,000,000 H. P. was 
used for manufacturing purposes alone in the United 
States, of which 21.7^ was derived from water and 
79.3^ from steam. As fuel becomes more expensive, 
however, the vast natural storehouses of water-power 
will surely be more and more drawn upon. It has 
been estimated that the rivers of the United States 
can supply some 200,000,000 H. P., and besides this 
there will always be available, while gravitation acts, 



128 Hydromechanics 

the power of the rising and falling tides. A simple 
calculation shows that a five-foot tide can furnish 
about 1800 H. P. from each square mile of ocean sur- 
face^ and doubtless the time will come when efficient 
machinery will be devised to utilize this source of 
power. 

(94) We will now consider briefly the methods by 
which the potential energy of a head of water is con- 
verted into useful work. 

However water may be used to do work, its available 
energy is always the product of its weight and head. 
A quantity of water, say a cubic foot, caught in the 
bucket of a water wheel at the top of a fall, and de- 
scending through the vertical distance h feet, does 
the work 62. 5h foot-pounds by the downward pressure 
of its weight. The same quantity of water, if allowed 
to enter a hydraulic press from the bottom of a 
reservoir in which the water stands h feet high will 
do the same amount of work, 62. 5h foot-pounds, by 
virtue of its pressure, which is 62.5h pounds per 
square foot. And, finally, if the cubic foot of water 
be allowed either to fall freely h feet from the top of 
the reservoir or to issue in a jet from its bottom, its 
energy, or the work it is capable of doing, is still the 

62.5v^ 
same, being — j — = 62.5h foot pounds. 

In the first case energy of position, measured by 
the potential head h, is converted directly into work, 
and in the second case energy of position, measured 

by the pressure head — = h, is converted directly 

into work. In both cases the work is done by the 
static pressure exerted by the water while it descends, 



An Elementary Treatise 129 

and any acquired kinetic energy is wasted. The only 
difference is that in the water wheel one cubic foot of 
water descends h feet, and in the press h cubic feet of 
water descend one foot, the result being 62. 5h foot- 
pounds of work done in each case. 

In the third case energy of position has been trans- 
formed into kinetic energy, measured by the velocity 

head -— = h, and this has then to be converted into 
2g 

work by utilizing the dynamic pressure caused by the 

impulse of the jet. 

Since there is no loss in the transmission of hydro- 
static pressure, and since pressure head can be con- 
verted into velocity head with hardly any loss, the 
choice of method of utilizing a given water power 
depends upon the cost and efficiency of the different 
kinds of motors and the practicability of their use 
under the particular conditions of each case. 

(95) When water under a head is used to do work 
by the direct action of its pressure, the energy is 
really only transmitted by the water and is not inher- 
ent in it. Of course it is elastically compressed and 
will do work if allowed to expand against a resistance, 
but this intrinsic energy, unlike that of a compressed 
gas, is always very small, however great the pres- 
sure, and a compressed liquid is only useful for trans- 
mitting energy, not as an independent source of it. 
It is convenient, however, to speak of the energy which 
water under pressure will transmit as the energy of 
the water itself, and in a sense this is true, since work 
can only be done when the water moves and so the 
amount of work done is really measured by the amount 
of water under pressure which is expended. 
9 



130 Hydeomechanics 

With this understanding, then, we say that the 
available energy per pound of water is h foot-pounds, 
where h is the pressure head in feet, and since p = 
hw, where p is the pressure per square foot and w 
the weight of a cubic foot of water, it is evident that 
the foot-pounds energy of water per cubic foot is its 
pounds pressure per square foot. 

This may be made more directly apparent by con- 
sidering the water to be doing work in a cylinder of 
one square foot cross section, when, evidently, the 
expenditure of one cubic foot of water under a pressure 
of p pounds per square foot will move the piston one 
foot, and so do p foot-pounds work. 

Thus the energy per cubic foot of water delivered 
at a pressure of 750 pounds per square inch is 750 X 
144 = 108,000 foot-pounds, and the use of a cubic 
foot of this water per minute in a motor of 80^ effi- 
ciency would furnish — = 2.62 H. P. 

(96) Water is sometimes supplied under pressure in 
mains for use for elevators and other domestic motors, 
and, as an example of such a case, suppose it is desired 
to deliver 2.5 cubic feet of water per second, at 600 
pounds pressure per square inch, through a 6" pipe 
2000 feet long, the distributing main being 18 feet 

4Q 40 

above the pumping plant. Thus v = — p- = = 

12.73 f. s., and-^= 2.53 feet; also the required pres- 
sure head is p^r— = 1382 feet. Therefore the 

62.5 

work which the pump has to do is equivalent to lifting 
2.5 cubic feet of water per second 1400 feet high and 



An Elemei^taey Treatise 131 

at the same time overcoming the resistance in the 
supply pipe. The losses of head due to this resistance 

are as follows : (1) h' = -r — ^ = 1.5 -— is used np in 
^ ^ 2gc^ 2g ^ 

imparting the velocity v to the water in the pipe; 

fl v' v' 

(2) h'' = ~Y • s~ = SO— is used np in overcoming 

v' 
friction; (3) h''' = m— , in which we will assume m = 

4, is lost in passing through the pump. The total 
work of the pump must be, therefore, 2.5 X 
62.5 [MOO + 2.53(1.5 + 80 + 4)] = 252,550 foot- 
pounds per second = 459 H. P., and the available 

• XI. ^- X -u X- • • 2-5 X 600 X 144 
power m the distnbutmg mam is ^^ = 

393 H. P., so that there has been a loss of nearly 
14.5/^ up to the distributing point, the loss from that 
point on depending, of course, upon the various cir- 
cumstances attending the use of the water by the 
consumers. 

Calculation will show that the use of a pipe a foot 
in diameter would reduce the required horse-power of 
the pump to 400, making the loss less than 2^. 
Doubling the diameter of a pipe, however, requires for 
equal strength to withstand internal pressure a four- 
fold weight, which means approximately a four times 
greater cost, and this has to be weighed against the 
greater economy in operation which would result. 

With such a system as just referred to an accumu- 
lator would be associated, to receive the surplus water 
at times of less than the average use, and to supply 
the deficiency at times of more than the average use. 

(97) In the transmission of water every loss of head 



132 Hydromechanics 

not directly resulting from a change of level means 
the dissipation of useful energy in heat, with no pos- 
sibility of its recovery, and the loss measured in foot- 
pounds is the product of the lost head and the weight 
of water delivered in pounds. It must be remembered, 

however, that the velocity head f-^ J enters as a factor 

into every loss of head, showing the great importance 
of keeping down the velocity when water-power is 
being transmitted, and also indicating that increased 
resistance to the flow, though it reduces the quantity 
of water discharged, may increase both energy and 
power. 

The partial closing of a valve in a pipe may cause a 
loss of head several hundred times the velocity head, 
but the resulting decrease in the velocity head itself 
may make the actual loss very small, and, in fact, this 
loss becomes zero when the infinite resistance to flow 
of a completely shut valve has made the velocity zero. 
If the outlet of a pipe be completely closed the velocity 
head vanishes, the pipe becomes a part of the reservoir, 
and the pressure head equals the hydrostatic head, and 
by altering the size of the outlet the energy of the issu- 
ing stream may be made anything from an infinites- 
imal fraction to practically 100;^ of that corresponding 
to the hydrostatic head. 

(98) Even with moderate velocities, the loss of head 
from friction is so great when water is carried through 
long pipes that for use as power it is almost always 
necessary to increase the energy of the jet by the use 
of an orifice of less size than the pipe itself, the loss 
of head from the contraction being reduced to almost 
nothing by a tapered mouthpiece or nozzle. 



An Elementaey Treatise 133 

To illustrate the effect of this, we will consider the 
case of a fire engine which pumps water through 250' of 
2"5 hose, the pressure at the engine being 125 pounds 
per square inch. If, then, the discharge be from the 

/ 2gh / 288 _ 

hose itself, we have v = / fT = 8 y i k , 94.— 

V 1.5 + ^ 

(taking f = .02) 26.9 f. s.; whence we see that of the 
total head of 288 feet, only 11.3 feet is utilized in 
velocity head, 276.7 feet head being used up in over- 
coming friction. 

But now suppose a nozzle of one-third the diameter 
of the hose to be applied at its end, and let v^ be 
the velocity of efflux, v being the velocity in the pipe. 

v' 
Then the losses, as before, are .5—- at entrance and 

2g 

fl v' v' 

^- r— = 24 — from friction, but the velocity head at the 

d 2g 2g ' ^ 

outlet is now ~- , and, since v^ = 9v, this equals -^ , 

and, moreover, the head to produce the velocity v^ in 

v' 
a tapered nozzle is 1.04-^ : so we have for the total 

(.5 + 24 -f 1.04 X 81)^ = h = 288; whence v = 

13 f. s.; and v^ = 9v = 117 f. s. The useful velocity 
head is now 213.9 feet and only 74.1 feet head has 
been lost. 

The quantity of water discharged per second is 
diminished from 0.917 in the first case to 0.425 cubic 
feet in the second case, but the power of the jet is 
increased from 648 to 5553 foot-pounds per second, or 
nearly nine fold. 



134 Hydromechanics 

(99) Neglecting loss of head at entrance to the pipe, 
and assuming a nozzle of perfect form, so that there 
is no loss at exit, we have 

I 2gh 

^ d "^d,* 
where d is pipe diameter and d^ nozzle diameter, and 

d' 
since the velocity of exit v^ = ^^ v, we have 

Now the work per second, or power of the jet is E = 

_ Vi^ Trd,^ N,' WTT / 2ghd,M^ \^ , , , ^. 

wQtt-^ w — - . ^ - =r--^ ^,° , 1- ; and by putting 
^ 2g 4 2g 8g \f Idi* + dV ^ 

the derivative of E with respect to dj equal to zero 

it will be found that E has a maximimi value when 

(35) d, = dyA. 

In the foregoing example, then, the maximum 
power would have been obtained by making the diam- 



di = di- 



eter of the nozzle ., _ . .y ^^-^ ^ x .02 X 250 

— -= = 0''95, instead of -'^- = 0'.'83; and by calcula- 
\/4o o 

tion will be found to be about 6310 foot-pounds per 

second, or nearly 11.5 H. P. 

(100) The distinction between the energy per unit 

weight of water, and the power, or energy per unit of 

time, of a stream, should be kept clearly in mind. The 

former is -r— = c/ h, (where c^ is the coefficient of 

velocity), and measures the efficiency of the jet, which 



An Elementaey Tkeatise 135 

efficiency, being the ratio of actual to theoretical 

c.^h 
energy, is e = — |- — = c/. The latter is the actual 

energy of the stream per unit of time, and so is E = 

wQ-p— = cwa^r-, where c is the coefficient of discharge 
2g 2g^ 

and a the orifice area. 

(101) When water flows from an orifice there must 
be a reaction of the jet equal and opposite to the 
force which gives it velocity. If, then, a be the area 
of an orifice with well-rounded entrance, under the 
head h, the velocity of efflux v is imparted in each 

second of time to a mass , and, since change of 

momentum in unit time is the measure of the force 

awv" 
which causes it, the reaction of the jet must be , 

which, taking the coefficient of velocity to be unity, 
equals 2awh . 

Hence the reaction of a jet, if we neglect loss of 
velocity by friction, is double the hydrostatic pressure 
on the orifice, and if, for example, the waters of 
Niagara could be received in the short arm of a ver- 
tical J shaped tube, they would balance a column twice 
as high as their 162-foot fall. 

(102) As an example of the utilization of the reac- 
tion of a stream of water for doing work, we will con- 
sider the " Jet Propeller.'^ 

The method of jet propulsion consists of taking 
water into a vessel and, by means of a pump, ejecting 
it sternwards, the reaction thus produced driving the 
vessel forwards. Let u be the velocity of the vessel 
and V the sternward velocity, relative to the vessel, 



136 Htdeomechanics 

imparted to W pounds of water by the pump each 
second. Then each second W pounds of water is 
accelerated from rest to an absolute velocity v — u, 

W 

and so the reaction, or propelling force, is — ^ (v — u), 

and the work of propulsion in foot-pounds per second is 

Wu 

(36) E = ^(v-u). 

As another way of looking at it, consider that the 
water, entering with the velocity u relative to the 
pump leaves it with the relative velocity v; whence 

W 

the whole useful work of the pump is ^(v^ — u^) . 

But the water leaves the vessel with the absolute 

W 

velocity v — u, and so carries off the energy^ (v — u)^ 

Therefore the work used in propulsion is the difference 

Wu 

of these, which is — (v — u) as before, 
g 
The efficiency of the apparatus is, of course, found 
by dividing the useful work by the total work, and is 

Wu, ^ 

—(V - u ) 

therefore -^ , so that we have 

(37) e==^ 

and this attains the value unity only when u = v, 
which could only happen if W were infinite. 

It will be observed that the foregoing is all based 
upon the supposition that the water enters with a 
sternward velocity u relative to the ship, and that 
the pump accelerates it to v. Actually the water in 



An Elementaky Tkeatise 137 

entering acquires part, and sometimes the entire for- 
ward velocity of the ship, in which latter case the 

Wu' Wv' 

energy — — is lost; the total work of the pump is -r — • ; 

the useful work is (v — u) as before: and the 

efiSciency is ^^ T ^^ = 2 — fl I which has a 

V 

maximum value 0.50 when u = - . 

The method by which the water is conducted 
through the vessel determines which of the foregoing 
values of e is most nearly correct. The value e = 
0.71 has been attained in an experimental boat, but 
the low eflBciency of the centrifugal piunp which forms 
part of the apparatus makes jet propulsion practically 
much less efficient than screw propu.lsion. 

Since the resistance is approximately f Au^, where 
f is a coefficient of friction and A is the wetted surface 

W 

of the vessel, we can put f Au^ = — (v — u) and to 

obtain an approximate value of the speed when we 
know W and v. 

Pkoblexs IX. 

(1) A pipe 6000' long and 1' in diameter leaves a 
reservoir 100' below its surface and runs horizontal 
for 1000'; the next 1000' inclines upwards till it is 27' 
above the first level; the next 1000' has a uniform 
downward slope of 1 in 20; the next 1000' of 1 in 10; 
and the last 2000' of 1 in 100. What is the velocity 
and what the pressure per square inch at the beginning 
of each 1000' length? If the outlet be closed what 



138 Hydromechanics 

does the pressure per square inch at the same points 
become ? 

(2) Water is flowing 16 f. s. in a pipe of 4" diameter; 
what is the loss of head due to a sudden change of 
diameter of the pipe to 8''? If the pressure head in 
the 4" pipe at the entrance to the 8'' pipe is 20', what 
is the pressure head in the 8" section? 

(3) If water is flowing 4 f . s. in a pipe of 8" diameter, 
what is the loss of head due to a sudden change of 
diameter of the pipe to 4"? If the pressure head in 
the 8" pipe is 20', what is it in the 4" pipe ? 

(4) If the power of the rising and falling tide 
could be all utilized, what horse-power could be 
obtained per mile surface at a place where the mean 
rise and fall was 3 feet ? 

(5) How much water at 400 pounds per square inch 
must be used in a motor of 75;^ efficiency to develop 
12 H. P. ? 

(6) It is required to furnish 100 cubic feet a minute 
of water at 500 pounds per square inch in a main 50 
feet above a pumping station. If the water is supplied 
through a mile of pipe of one foot diameter, what 
must the power of the pump be, and what is the per- 
centage lost in delivering the water? 

(7) What horse-power must a pump have to deliver 
2,000,000 gallons of water per day through an 8" pipe 
1500 feet long at a station 220 feet above the pump? 
What part of the power is lost in overcoming resist- 
ances ? 

(8) What should the diameter of nozzle be for 
maximum power in the case of a 6" pipe, 400 feet long, 
with a total head of 200 feet, and what is that power ? 

(9) What horse-power is furnished by discharge 



An Elementary Treatise 139 

through a o" pipe 800 feet long^ with a total head of 
■iOO feet, and what will it be if a nozzle 0'.'82 in diam- 
eter be used ? 

(10) What diameter of nozzle should be used to 
get the most power from a 300-foot head discharging 
through 1300 feet of 1-foot pipe, and what horse- 
power will be furnished ? 

(11) A pump of 100 effective H. P. discharges 
directly into a stand pipe of 20 feet diameter from 
which 18,000 cubic feet per hour of water is being 
dra^vn. How long will it take to raise the water level 
from 80 to 100 feet ? 

(12) If a jet is an inch in diameter, what must its 
discharge be to make the reaction 200 pounds? 

(13) What is the reaction of the jet from a nozzle 
2" in diameter with a pressure head of 200 feet at the 
nozzle entrance? 

(14) The Magara Power Company has 21 double 
turbines, each developing 5000 H. P. under a head of 
water of 136'. If their average efficiency is. 0.75, how 
many cubic feet of water per second must be drawn 
from the falls for their use ? 

(15) If the resistance in pounds to the motion of a 
vessel is 4 times the square of her speed in f . s., what 
will be her speed when propelled by the reaction of a 
jet from a 2'' nozzle under a pressure of 150 pounds 
per square inch? What is the efficiency of this 
propeller ? 

(16) What must be the effective H. P. of the pumps 
to give 8 knots speed to a vessel whose resistance is 
5v^ pounds at a speed of v f. s. by means of a jet from 
a 2" nozzle, and how much water must be pumped 
through the vessel per second? 



140 Hydromechanics 

(17) An experimental torpedo-boat of 13 tons dis- 
placement, built by Thorneyeroft in 1881, was given 
a speed of 12.6 knots by the discharge of a ton of 
water per second at 37.25 f. s. The I. H. P. was 170 
and the efficiency of the engine 0.77, what was the 
efficiency of the pump; what was the efficiency of the 
jet propeller; and what does the latter value indicate 
as to the method of taking in the water? 

(18) The Water-witch, an experimental jet propeller 
of 1160 tons made 9.3 knots with 760 H. P. Her water 
supply pipes were so arranged that the water prac- 
tically attained the ship^s speed before being expelled 
rearward. The efficiency of her engine was 0.70; that 
of her pumps 0.45; and the escape orifices aggregated 
5J square feet. How much water did she discharge 
per second; at what velocity; and what was the total 
efficiency of her propelling machinery? 



An" Elementaey Teeatise 



141 



CHAPTEE X. 

Reaction and Impulse — Water Motors. 

(103) The so-called Barker^s Mill is another example 
of a pure reaction motor. This consists of a hollow 
drum kept full of water under pressure, and having 
hollow radial arms from the outer ends of which the 
water issues in jets tangentially 
to the motion of rotation caused 
by their reaction (Fig. 27). 
Suppose the pressure head h to 
be maintained in the drum by 
the influx of water through a 
pipe so large in relation to the 
efflux that the velocity of ap- 
proach may be neglected, and 
let u be the linear velocity of 
the orifices, v the relative veloc- 
ity of efflux, and W the weight of water discharged 
per second. Then the total energy supplied per second 
is Wh; and the effluent water carries off in the same 

W 

time the energy—— (v — u)-; and the difference, 




FiG.27. 



^g 



W h - 



(Y - U)' 



2g 
wheel. But v - 



must be the work done on the 



V^gh -\- u^ , since the total head on 
the orifices is the static head (h) plus the head due 

to rotation ( -,r— ). 

W 

— (uv — u^) or. 



Therefore the useful work is 



143 HYDEOMECHA.NICS 



(38) E=^(v-ii). 

This seems to show that the reaction of the jets is 

W 

— (v — u), whereas the absolute change in velocity, 

being from u in one direction to v — n in the other, is 

Wv 

really v, and so the reaction must be . The 

explanation of this anomaly is that there is a constant 

force opposed to the reaction of the jets, due to the 

fact that each second the weight of water W, entering 

the drum without velocity, must travel to the orifices, 

and in so doing acquire the velocity u and the energy 

Wu=^ 

-^ — . But the distance it travels per second in the 

line of action of the force giving it rotation is only-. , 

and so the force constantly required to keep up the 

Wu^ u Wu 

rotation of the water is -^ '- -^= , and this 

2g ^ g 

Wv 

must be subtracted from the reaction of the lets 

g 
W 

to get the effective rotating force (v — u). 

(104) The efficiency of the Barker's Mill is the use- 
ful work ^^ divided by the total work 

g 
W 
Wh = -— (v^ — u^), and is therefore the same as that 

of the jet propeller, 

(39) e= ^" 



V + u 



Since v = \/2gh. -j- u^ , u can never be as great as v, 
but the efficiency approaches unity as u increases in 



An Elementary Treatise 143 

value, and coiild u = oo we would have v = oo also 
and so e = 1. The weight of water discharged per 
second, W = wav (a = orifice area), would also be 

infinite; the rotating pressure, (v — u) = 

wav , 



— (v — V"^^ — ^E^)} evaluated for v = oo will be 

found to be wah, the same as the static pressure due to 
the head alone; and the work of the wheel, being u 
times this, would be infinite. 

In practice there is some reduction in the value of 
V from friction, and as speed increases the resistance 
of the air becomes important, so that the simple 
reaction wheel is no longer used as an hydraulic motor 
on account of its low efficiency. 

(105) Just as it requires pressure to give water 
velocit}^, so pressure is exerted by moving water against 
any surface which either changes its direction or 
checks its velocity. Such pressures are called 
'^ dynamic pressures,^^ and, like other impulsive forces, 
they are measured by the change in momentum which 
they produce in unit time. 

If the jets issuing from a Barker^s Mill are caused 
to impinge on surfaces rigidly secured to the mill 
itself, it will be found that when those surfaces are 
planes normal to the jets there is no rotation of the 
mill; when they are convex to the jets rotation in 
the usual direction occurs; and when they are concave 
to the jets there is rotation in the opposite direction. 
This shows that the dynamic pressure is greater the 
greater the angle at which the impinging jet is 
deflected. 

When a surface completely destroys the velocity of 



144 



Hydeomechanics 



a jet in its original direction^, all the water being 
deflected 90°, the dynamic pressure in that original 
direction is called the " impulse of the jet," and, as 
shown by the experiment just referred to, equals the 

Wv 

reaction of the let . 

g 

(106) We shall now show that the total dynamic 
pressure in a given direction caused by a stream which 
impinges upon a smooth surface, passes over it, and 
then leaves it, is measured by the difference between 
the components, in the given direction, of the im- 
pulse of the approaching and the reaction of the 
departing stream, and is independent of the form of 
the surface. 

Let AB, Fig. 28, be a horizontal section of a smooth 
vertical vane, and suppose that a weight of water W 
per second enters it with uniform velocity v at A and 

leaves it at B. It is then 



Fio.28. 




required to find the total 
dynamic pressure on the 
vane in the direction in- 
dicated by the arrows 
marked u. The stream is 
to be prevented from 
spreading, so that all the 
water will leave at B 
at the same velocity as it 
had at A, and at first we 
will suppose that it en- 
ters tangentially at A. Calling s the length of the 
path AB, p its radius of curvature at any point P, 
and 6 the angle that radius makes with a normal to 
the direction u, the mass of water on the vane is 



Ax Elemei^taey Tkeatise 145 

Ws 
Ws ^d^ Wpdd 



, and the mass on the element odd is 
gv 



There beinff no friction, the only 
gv s gv ^ ^ J 

force acting on this mass is the centrifugal force 

W^d^ v^ Wvd^ ^ ,, . , . , . 

• — = , and this mnst be the pressure on 

gN p g 

the element p&d and acting normally to it. The com- 

Wv 

ponent of this pressure in the direction u is sin 

Odd, and therefore the total pressure in that direction 

f^Wvsin Odd Wv ^ , .. T.„ ,, 

- (cos a^ — COS y5). If now the 



"I 



stream approaches^ not tangentially, but at the angle 
a with u, we must suppose it to be guided by a rim 
(as shown at (b). Fig. 28) in order that it may all 
enter the surface, and, however abrupt the change of 
direction at K may be, we can consider this rim as 
forming part of the vane and extend the integration 
to include it. So we have for the total pressure in 
the given direction, in all cases, 

Wv 

(40) F, = ^^ (cos a - cos /?) . 

Wv 

And, since is the impulse of the approaching 

Wv 

stream, and cos a. its component in the given 

Wv 

direction, and since, similarly, cos /5 is the com- 

ponent in the given direction of the reaction of the 
departing stream, our proposition is proved. 

Of course, when /5 exceeds 90°, as in Fig. 28, its 
cosine is negative, and so the reaction of the departing 
is added to the impulse of the approaching stream. 



146 



Hydkomechanics 



(107) If the stream does not enter nearly tan- 
gentially, there will be some loss of energy caused by 
eddies, foam, and changes of section, but neglecting 
this, the results of experiments fully accord with the 
foregoing theoretical deduction. For consider a sur- 
face of revolution with an impinging jet as shown at 
(a), Fig. 29. Then, since each element of the stream 
enters tangentially, our mathematical deduction ap- 
plies rigorously, and the dynamic pressure in the direc- 

Wv Wv 

tion of the let is (cos 0° — cos 90°) = . 

But a simple weighing arrangement, when a flat plate 




Fig. 2- 9. 



is used, as shown at (b), gives this same result, within 
about 4^, thus Justifying our assumption that provided 
the stream enters the surface it is the direction of the 
approaching jet, and not that of the surface, which 
determines the impulse. 

If then we apply our formula (40) to determine the 
pressure in the direction of the jet with a surface 
such as is shown at (c), we have a = and ,3 = 180°, 

2Wv 

whence F^ 



g 



With the arrangement shown at 



(d), where the jet suffers a double reversal of direction. 



An Elementary Treatise 147 

4Wv 

we should have theoretically F^ = , and experi- 
mentally the value '-^ has been found, the loss 

g 
from friction, etc., being in this case naturally large. 

(108) The total action, then, of a jet impinging 
upon any smooth surface is equivalent to the action 
of two forces, one the impulse of the approaching 
stream acting at the entrance, the other the reaction 
of the departing stream acting at the exit. 

If we wish to determine the amount and direction 
of the resultant of the total action of a jet on a 
surface, we have only to combine the aforesaid impulse 
and reaction as we would combine any two forces to 
find their resultant. 

If we wish to determine the pressure in any direction 
caused by the action of the jet upon the surface, we 
have only to take the algebraic sum of the components, 
in that direction, of the aforesaid impulse and reaction. 

In fact, the conclusion we have just come to amounts 
to nothing more than this : that the change of momen- 
tum in any direction per second measures the force 
acting in that direction, — for the momentum per 

Wv 

second in the direction u was originally cos a, 

Wv 

and finally it is cos ,3 in the same direction, and 

Wv 

so there must have been a force (cos a — cos /3) 

acting on the water in the direction opposite to u, and, 
since the same change of momentum takes place in 
each second, there must be a continuous pressure 

AVv 

(cos a — cos /S) acting to produce it. 



148 



Hydromechanics 




Fig. 30. 



(109) In order, however, that work may be done by 
the pressure on a vane, the vane must move. Let 

u be the uniform linear 
velocity of the vane caus- 
ed by the dynamic pres- 
sure Fu in the same 
direction. Then evi- 
dently the pressure de- 
»^ pends upon the relative 
velocity of the water in 
j^^^r... contact with the surface, 
and formula (40), de- 
duced for the case of a stationary vane, will apply to 
this case provided merely we use the relative velocity 
V = v^ + u^ — 2uv cos a instead of the absolute 
velocity v (Fig. 24). 

WV 

Therefore we have, in this case, F„ = (cos ^ — 

cos /5), and since V sin w =i \ sin a, V^ cos^ <p = 
V^ — v^ sin^ a = (v cos a — u)^, we get 

W 

(41) Fu = (v cos a — u — V cos /5) . 

Moreover, as Fig. 30 shows, u + V cos /S = v^ cos d 
and so we have also 



(42) 



W 

Fu = (V cos a 

g 



Vi cos 0) , 



Formula (41) shows that the total pressure in the 
direction u is the difference between the impulse and 
the reaction due to the relative velocities in the given 
direction of the approaching and departing streams; 
while formula (42) shows that the total pressure is 
also equal to the difference between the impulse and 



An" Elemextaey Teeatise 14:9 

the reaction due to the absolute velocities in the given 
direction of the approaching and departing streams. 

(110) The work done on the vane per second;, then^, 
is either of the expressions — 

(43) E =-^^^Cv cos a - u — V cos /5) 

= (V cos a — Vj cos d) , 

and a consideration of the energies of the entering and 
departing streams will give the same results. For the 
absolute velocity of the jet before it touches the vane is 
V, and when it leaves the vane is v^ = V ^" + ii^ + 2u Vcos ,3, 
and so the work done on the vane, neglecting friction, 
must be the difference of the corresponding energies, 

W W Wu 

or-— (v^ - Vi^) =^(v^ _ V^ - u^ - 2uYcos ^5) =-^' 

(v cos a — u — V COS /?) as before (since V^ = u^ + v^ 
— nv COS a, or uv cos a = v^ + u'^ — V^) . 

Either of the two foregoing methods may be used 
to determine the power of a hydraulic motor which 
is actuated by the dynamic pressure of an impinging 
stream, and though we have assumed all parts of our 
vane to have the same linear velocity, which is not the 
case in practice, it is still always true that the work 
done, neglecting frictional losses, is the difference be- 
tween the energy of the entering and that of the de- 
parting stream; and is, equally, the difference of the 
works done by the impulse of the entering and the 
reaction of the leaving stream. It is either of the 
expressions, 

AV W 

(44) E = — - (v^ — Vj^) = — (uv cos a — UjVj cos e) 

in which v and v^ are the absolute velocities of the 



150 Hydromechanics 

stream in entering and leaving; n and u^ the velocities 
of the two parts of the vane where the stream enters 
and leaves it; and « and 6 the angles respectively 
between v and u and between v^ and n^ . 

(Ill) Water wheels driven by the impulse of a jet 
consist of a number of vanes which turn about an axis, 
and come successively in front of the jet so that such 
portions of the water as do not strike a vane just 
leaving the jet are utilized by another vane just 
coming into position, and we may thus properly con- 
sider such a motor as consisting of a single vane against 
which the entire weight of water discharged per 
second (W) is being constantly delivered. 

Considering the expression for the work of a wheel, 

W 

- — (v^ — Vj^), it is evident that for maximum efficiency 

there must be as little loss of energy from foam and 
friction as possible, and the absolute velocity of exit 
must be zero, which conditions are summed up in the 
statement that " the water must enter the wheel with- 
out shock and leave without velocity/^ 

The first condition is met by so shaping the vanes 
that the relative motion of the entering stream is 
tangential to the surface it first impinges upon, which 

requires the relation — = r- (Fig. 30), a 

' V sm ^ ^ * 

being called the approach angle and <p the entrance 

angle. 

The second condition requires the exit angle /? to 

be 180°, and also that v^ shall equal u^ , which would 

make V = u and so 2^ = a ; but /? cannot have this 

value, since if it did the water would not leave the 

wheel, and so the best that can be done practically is 



An- Elementaey Treatise 151 



to make /S from 150° to 175° and o. = ~^ This 

Z 

gives V = 2ii cos «, and v^ = 2u^ cos — . But the efl&- 
ciency of the wheel is ^ ^^ = 1 \ , therefore 



Y 

3g 



(45) e = l- <"°^- =l- '-'°°^ 2- 

U^ COS'^ a r^ COS^ a 

in which r and r^ are the radii of the inner and outer 
edges of the vane. 

Formula (45) indicates that « should be as small as 
possible, but it must be large enough to make the 
water enter the wheel, and its size is less important 
than is that of /?. 

The equation v ^ 2u cos «, which results from the 

condition c? =z 2a , shows that u should equal 

^ 2 cos a 

for the maximum efficiency, so that the smaller « the 

y 

nearer u should approach— . 

(112) As an example of the foregoing formulae, we 
will determine the proper speed, the power, and the 
efficiency of a wheel, having the following character- 
istics: r = 3'; T, = 4:'; a = 30°; c^ = 60° ; /5=150°; 
and driven by the jet from a fixed nozzle, . v being 
100 f. s. and W = 125 pounds. Then to avoid shock 

y 

at entrance u = = 57.7 f. s., and so the wheel 

2 cos a 

should be adjusted to make 3.6 turns a second, or 216 

4u 
turns a minute. But, since Uj^ = -^ = 76.9 f. s., v^ = 2u^ 



152 Htdromechanics 



cos -^ = 39.8 f. s. Therefore the work per second of the 

w 

wheel is ^ (v' - v^^ = 16,426 foot-pounds = 29.9 H.P.; 

and the efficiency is 1 j- = 0.841. 

(113) A very simple, and yet a very efficient water 
motor is a vertical impulse wheel specially designed for 
use with comparatively small quantities of water under 
great heads, and known in California, where it is much 
used, as a " hurdy-gurdy '' wheel. It is an iron wheel, 
of diameter from 1 to 6 feet, according to the power, 
carrying on its outer rim a number of cups into which 
the jet from a fixed nozzle plays as they successively 
come into position in front of it. Each cup has a 
projecting sharp edge or rib across its center, to split 
the impinging jet, and is so formed that the direction 
of the stream is almost completely reversed after 
impact. It will easily be seen that the efficiency of 
such a wheel is greatest when the speed of the cups 
is half the velocity of the jet, since in that case the 
absolute velocity of the water after impact, provided 
the shape of the cups was such as to exactly reverse 
the motion of the water, would be zero, and the effi- 
ciency would then be unity, all the energy of the jet 
being transmitted to the wheel. In actual practice 
these wheels attain an efficiency of 0.85 and sometimes 
even more. 

Peoblems X. 

(1) What is the efficiency of a Barker's mill when 
the linear velocity of the orifices (u) is half the velocity 
of efflux (v), and what is the power under such condi- 



An Elementary Treatise 153 

tions if the head on the wheel is 25' and the orifice 
area is 4 square inches ? 

(2) The linear velocity of the orifices of a Barker^s 
mill is ^/2gh, where h is the head on the mill. What 
is the power and the efiiciency ? 

(3) If the velocity of efflux of a Barker^s mill is only 
0.94 its theoretical value (Cg = 0.94), show that the 
most advantageous linear velocity of the orifices is 
nearly u = V^gh and that the efficiency is then e = 
0.66. 

(4) A 1" nozzle delivers a cubic foot a second. What 
dynamic pressure will it produce against a plane nor- 
mal to the jet, and what will the pressure be if the 
plane is inclined 30°, 45°, and finally 60°, supposing 
all the water to be guided in the direction of maximum 
deflection ? 

(5) A stream of 64 square inches section and with 
a velocity of 40 f . s. impinges on the vertex of a fixed 
cone of 45° semi-vertical angle and in the direction 
of its axis. What is the force acting on the cone? 
If a sphere replaces the cone, what is the action ? 

(6) Water flows at the rate of 2 cubic feet a second 
through a flexible hose of 3'' diameter. If the hose is 
lashed to two eye bolts so that the parts leading from 
them to the bend in the hose make an angle of 60°, 
the rest of the hose being free, what is the pull on 
each bolt, and what is the whole dynamic pressure on 
the hose between the bolts ? 

(7) Water is flowing at the rate of 8 f . s. through a 
3" pipe. The sudden closing of a valve stops the water 
in a 200-foot length of the pipe in 0.10 seconds. What 
is the increase of pressure near the valve? 



154 Hydromechanics 

(8) Two cubic feet of water moving 20 f. s. im- 
pinge on a stationary vane and are thereby deflected 
60°. What is the pressure on the vane in the direction 
of and normal to the jet ? 

(9) If a jet impinges normally on a succession of 
plane surfaces which move away in the direction of the 
impulse, what must the relative velocity of the vanes 
be for maximum efficiency and what is that efficiency ? 

(10) If a series of curved vanes are successively im- 
pinged upon by a jet and move away in the direction 
of the impulse, how should they be shaped, and with 
what relative speed should they move to secure maxi- 
mum efficiency, and what is that efficiency? 

(11) A 2'' nozzle, under a head of 80', and with a 
coefficient of velocity 0.94, delivers its jet normally 
against a series of small plane vanes moving in a cir- 
cumference of 16' diameter. How fast should the 
wheel be made to turn and what then is its power ? 

(12) A 1" nozzle, under a 100' head, coefficient of 
velocity 0.98, delivers its jet against a series of cups 
moving in a circle of 2' diameter and so shaped that 
they reverse the direction of the impinging water. 
How many revolutions should the wheel make, and 
what would its horse-power then be? 

(13) A ^^ hurdy-gurdy " wheel of 6' diameter is 
actuated by the jet from a 1''89 nozzle under a head 
of 386', the water being conveyed to the nozzle through 
6900' of 22" pipe. If the efficiency of the wheel was 
0.87, what was its H. P. ? 

(14) Compute the power and efficiency of a water 
wheel of the following characteristics: r = 2'; r^ = 3'; 
a = 30° ; ^ = 60° ; /3 = 120°, when actuated by 2.2 



Ax Elemextary Teeatise 155 

cubic feet of water per second at 100 f . s., and when the 
speed of the wheel is 184 revolutions a minute. 

(15) Given r = 2'; r^ = 3'; a = 45°; cp == dO"" ; 
and /5 = 150°; what is the work of a water wheel and 
what is its efficiencj when actuated by 2.2 cubic feet 
of water per second at 100 f. s.; the speed of the wheel 
being 337.5 turns a minute? 



156 Hydromechanics 



CHAPTEE XI. 
Hydraulic Machines — Work of Expanding Gases. 

(114) When the energy of a natural head of water 
is to be transformed into useful work, the water is 
collected in a reservoir, canal, or head race; is thence 
conveyed through a pipe, penstock, or flume to the 
hydraulic motor; and, after passing through or over 
the latter, issues into the lower level, or tail race. 

The hydraulic motor generally consists of a wheel 
which is caused to revolve either by the descending 
weight of the water or by its impulse, and such a 
motor is called a " water-wheel ^^ when the water enters 
it at only a part of its circumference, and a " turbine '^ 
when the water enters around its entire circumference. 

(115) The " overshot " water-wheel is commonly 
used for falls of considerable height. Its motive force 
is principally the direct action of the weight of the 
water which is received in its buckets near the top of 
the fall and descends in them until emptied in the 
tail race. These wheels are sometimes 60' and more 
in diameter, and have an efficiency from .70 to .90, 
one of their special advantages being that when the 
water supply runs low their efficiency is highest on 
account of the buckets being only partly filled and so 
the loss of water by spilling before the bottom of the 
fall is reached being a minimum. 

The " undershot " water wheel is operated by the 
impulse of water which flows nearly horizontally be- 
neath it and acts upon its radial vanes. These wheels 



Ajst Elementaey Teeatise 157 

commoiily have an efficiency from .20 to .40;, and 
which is least when they are turned by the current 
of an unlimited stream, as in the case of a floating 
mill anchored in a river. When the vanes are curved 
so as to make the water leave with a less absolute 
velocity than that of the wheel itself, the undershot 
wheel may have an efficiency as high as 0.60. 

" Breast " wheels operate partly by the weight and 
partly by the impulse of the water. In them the 
buckets of the overshot wheel are replaced by vanes 
which move in a masonry channel partty surrounding 
the wheel. There is, of course, considerable leakage 
past the vanes, which cannot fit the channel closely, 
and the efficiency of the breast wheel varies from .50 
to .80. These wheels must be of greater diameter 
than the height of the fall, and so are only used for 
comparatively small falls. 

Other water-wheels, which are actuated by the im- 
pulse of a jet directed against their vanes, are known 
as " impulse " wheels, " vertical,^^ or " horizontaV^ 
according to the position of the wheel. Horizontal 
impulse wheels, further, are " outward flow," or " in- 
ward flow," wheels, according to whether the horizontal 
jet actuating them enters the wheel at its inner and 
leaves at its outer circumference, or vice versa. There 
are also horizontal " downward flow " wheels, actuated 
by one or more jets which pass down over the vanes 
without either approaching or receding from the 
wheeFs axis. These impulse wheels have efficiencies 
from .75 to .85. 

(116) Turbines, as already stated, are wheels in 
which the water enters around the entire circumfer- 
ence so that all the vanes are simultaneously acted 



158 Hydromechanics 

upon. They are usually horizontal, and may be " out- 
ward/^ " inward/^ or " downward " flow turbines, 
according to how the water passes through them. In 
all turbines the water passes through fixed curved 
guide passages which give it the proper direction to 
make it enter the wheel passages without shock, and 
the latter are also curved so as to cause the water 
leaving the turbine to have as little remaining energy 
as possible. 

Turbines are also divided into the two classes of 
" impulse '' and " reaction '^ turbines, though any tur- 
bine will act either as a reaction or as an impulse 
turbine according to whether the supply of water 
completely fills the wheel passages or not. If the 
wheel passages are only partly filled with water the 
action is entirely impulsive, but when the water fills 
them it transmits more or less static pressure into 
the wheel. The theory of impulse turbines is the same 
as that of impulse wheels, and the shape of their guide 
and wheel passages is determined as set forth in the 
last chapter. In reaction turbines, however, where 
the discharge of water, just as in the Barker's mill, 
varies with the speed of rotation and with the cross 
sectional area of the wheel passages, the problems of 
design and of the theoretical determination of effi- 
ciency are more difficult, and will not here be entered 
upon. 

The efficiency of well designed turbines, as experi- 
mentally measured, is from .70 to .80. 

(117) Every hydraulic motor may be made to raise 
water, or act as a pump, by the application of power 
to give it a reversed motion. The reversed overshot 
water-wheel, which dips up water in its buckets and 



An Elementaey Teeatise 159 

raises it to a height somewhat less than the diameter 
of the wheel is known as a " Chinese wheel." A 
reversed breast wheel is known as a " scoop " or 
" flash " wheel. The reversed inward flow turbine is 
the well-known ^' centrifngal '' pump. 

The different conditions, however, under which 
motors and pumps work sometimes makes considerable 
differences in design necessary for efficiency in the two 
cases. 

(118) As a rule water pressure engines are operated 
by an artificial head produced by steam pumps working 
in connection with an accumulator, though sometimes 
they are used with natural heads of water. They are 
of various forms, single or double acting, and with 
one or more cylinders, a not uncommon form being 
three single acting oscillating cylinders, inclined at 
120° to each other, and driving the same crank; but 
in all cases the principle is the same, and it may be 
fully illustrated by a consideration of the simple ram, 
such, for example, as is used to lift a weight by direct 
action to a height equal to the stroke. 

Let po be the pressure in the accumulator, p that 
in the working cylinder, A the piston area, and V the 
piston speed when lifting a weight W. Then the use- 
ful work per second is WV = ApA^, and p is less than 
Po by an amount which measures the sum of all the 
losses of head between the accumulator and the piston. 
But the total loss of head can be expressed in the 

V p — p V^^ 

form k-r— , and so we have -^ = k r- , or p = 

2g ' w 2g' ^ 

Po — kw ^ ; whence the useful work per second is — 
(46) E = WV = a(p„V - kw g ) . 



160 Hydromechanics 

If the weight being lifted is reduced, diminishing the 
useful work being done^, V increases, and so too does 

the loss of head k^; and, in the limiting condition 



W = 0, V reaches the maximum value J -^ , and 
' T wk 

the loss of head is — or the total head due to the 
w 

accumulator pressure. 

Thus a water pressure engine is automatically self- 
regulating as regards speed, the hydraulic resistance 
increasing as the load is diminished. 

Again, solving equation (46) for V, and putting Ap 
for W, we see that for any given load there is a certain 
" speed of steady motion " given by the equation 



(47) V = ^ ^g(P'-P) 
T kw 



and, since k can be increased at will by more and 
more nearly closing the valve between the accumulator 
and the cylinder, the speed of a water pressure engine 
can be regulated at pleasure. 

Furthermore, by differentiating (46) we find that 

the power is a maximum when V = a/ ^ v^", which is 

-= times the speed under no load, so that the power 
Y o 

of a water pressure engine supplied by a given pipe is 
greatest when one-third the total head is lost in over- 
coming frictional and hydraulic resistances. 

(119) It should be noted that the piston must start 
from rest and so there must be accelerated motion 
during the first part of each stroke, which reduces 
the effective pressure in the cylinder. Moreover all 



An Elementary Treatise 



161 



the water in the cylinder and supply pipes^ as well as 
the piston and the weight it lifts, must be accelerated, 
and, since the pipes are smaller than the cylinder, the 
water in them must have a correspondingly greater 
acceleration than that of the piston, so that there 
must be a material reduction in the effective pressure 
while the inertia is being overcome, and so a consider- 
able time may elapse before steady motion is attained. 




Fi&.31 



So, too, if the motion is a reciprocating one, there is a 
similar retardation and increase of effective pressure 
during the latter part of each stroke. Thus a water 
pressure engine is equivalent to a machine with very 
heavy moving parts, and is liable to considerable varia- 
tion of pressure. This, together with the incompres- 
sibility of the water, renders it necessary to apply 
11 



162 Hydromechanics 

relief valves or air chambers to such a system to pre- 
vent excessive pressures. 

(120) Pumps of common form are in principle 
merely the reversal of water pressure engines, the 
main difference being that in the former the valves 
open and close automatically by the action of the water 
while in the latter the valves are either worked by 
hand or by a connection with some moving part of the 
machine. 

Fig. 31 shows a double-acting force pump with an 
air chamber to maintain a continuous delivery, and 
which is actuated by a steam cylinder (not shown) 
whose piston rod is an extension of that of the pump 
itself. 

(121) The hydraulic cylinder is often used as a 
brake, or means of bringing a moving weight to rest 
without shock. In its simplest form the hydraulic 
brake is a cylinder full of liquid fitted with a piston 
pierced with one or more holes, so that motion of one 
part with reference to the other forces the liquid 
through the holes from one side of the cylinder to 
the other, thereby producing a pressure which is evi- 
dently proportional to the square of the velocity of 
relative movement. Such a device by itself would 
never bring a moving weight to rest, and moreover 
has the disadvantage of causing a maximum pressure, 
when the velocity is greatest, some four or five times 
the uniform pressure which would produce the same 
result. It is therefore now usual to make the escape 
openings from one side of the piston to the other of 
varying cross section, so that the escape area is great- 
est at the beginning of the motion to be checked and 
is gradually diminished in such a way as to make the 



An Elementary Treatise 163 

retarding pressure in the cylinder Tiniform. This is 
readily accomplished by having the openings in the 
form of shallow grooves in the inner surface of the 
cylinder, either the depth or the width of the grooves 
being gradually reduced from one end of the cylinder 
to the other. 

(123) Thus far the work done by liquids only has 
been considered, and something must be said in regard 
to the energy of gases. 

It has been pointed out that a confined gas possesses 
a store of energy within itself and, unlike a liquid, 
performs useful work by expanding. We will now 
show that this internal energy is, for any given mass 
of any gas, a function of its temperature alone, and 
does not depend upon either its volume or its pressure. 

As stated in Chapter I, experiment has shown that 
for all gases, not near the state of liquifaction, the 
product of the pressure and volume of any given mass 
divided by its absolute temperature is constant. When 
unit mass is taken and this relation expressed in the 
form 

(48) pv = RT. 
it is called, the characteristic equation of the gas con- 
sidered. Thus, since a cubic foot of air at 32° F. and 
14.7 pounds per square inch pressure weighs .0807 

pounds, K = ^»=jg^3^^^^- 53.3, and for air, 

py = 53.3 T, in which p is the pressure in pounds 
per square foot and v the volume in cubic feet of one 
pound of air at an absolute temperature of T° F. and, 
too, E for any other gas is evidently found by dividing 
53.3 by the specific gravity of the gas, thus being for 
coal gas, the density of which is .43 that of air, 123.95. 



164 Hydkomechanics 

(123) If now we suppose the pressure p to remain 
constant while the volume changes dv, (48) shows that 

the temperature will change K^ , and so the gas must 

have received the quantity of heat —^ — when c^ is 

the specific heat of the gas at constant pressure. , 

Also, supposing the volume v to remain unchanged 
while the pressure varies dp, (48) shows a change of 

temperature —^ and consequently the absorption of 

a quantity of heat -p , where c is the specific heat 

at constant volume. 

Therefore, when both volume and pressure change, 
the quantity of heat received by the gas is 

(49) dq = -^ (c'pdv -|- cvdp). 

Now, diiferentiating (48), we get 

(50) pdv + vdp = EdT 
and, by eliminating dp from (49) and (50), 

(51) dq = cdT + ^-^^pdv. 

(124) Considering (51), the quantity pdv is the 
external work done by the gas in changing its volume 
by the amount dv, for the work done by each element 
of the surface enclosing the gas is the product of its 
area, the pressure, and the distance it moves normal 
to itself, and, the pressure being constant over the 
entire surface, the sum of the works done by all the 
elements of the surface is the product of the pressure 
and the change of volume. Now, integrating (51), and 
calling Tq and T^ the initial and final temperatures and 



An" Eleme^s^tary Teeatise 165 

W the total external work done by the elastic force of 
the gas, we have 

(52) q = c(T,-T„)+-5^W, 

which gives the quantity of heat absorbed by nnit mass 
of a gas when it changes temperature from Tq to T^ 
while doing the work W, and by which, if the foot, the 
pound, and the degree Fahr. are the units used, q will 
be given in British thermal units. 

If in (52) we make T^ = To , implying that the gas 
has the same temperature in its final as in its initial 

state, we sret W = -, q, so that — is what is 

' ° — C 0— c 

known as the mechanical equivalent of heat and 
usually designated by J. Thus in the case of air, tak- 
ing E = 53.3 as already calculated, c' = .2375 as 

c' 
found by Eegnault, and y = — = 1.408, we get J = 

775, a near approximation to the value 779 which the 
best recent experiments assign to it. It should be 
observed, therefore, that, while the value of E is differ- 
ent for different gases, the theory of the conservation 

of energy requires the quantity —, to be inde- 
pendent of the nature of the body which serves as 
intermediary in the conversion of heat into mechan- 
ical work or of work into heat. 

If, then, no heat is either gained or lost by the gas, 
so that q = 0, we have W = cJ(To — T^), showing 
that the external work which a gas can do depends 
only upon its change of temperature, and has a maxi- 
mum possible value cJTq, which therefore is the in- 
ternal energy of unit mass of gas at temperature To . 



166 Htdromechanics 

(125) In isothermal expansion, then, the work done 
by any gas exactly equals the heat imparted to it to 
maintain its temperature, and in adiabatic expansion 
the work done is the product of the change of tem- 
perature by the specific heat of constant volume, in 
both cases the work being in heat units and to be 
multiplied by 779 to get its value in foot-pounds. 

Frequently, however, we desire to determine the 
work done by a gas in changing volume when the tem- 
perature is not part of the given data. If the change 
is isothermal we have then only to put dT = o in 
(50) and by integrating we get pv = k ^ p^v^ , whence 
the work done in expanding from v^ to V2 is 

fv* rva dv V p 

pdv = PiV,— = p,v, loge-^= PiV, loge-^, so 
Vi Jvi V Vj P2 

that the work done by any gas in changing volume 
isothermally is given by 

(54) W = p,y , log. ^^ = p,v, log, £i . 

If, on the other hand, the change is adiabatic, we 

get, by putting dq = in (49), -^ + -^ • ^ = o, 

or by integration, log p + — log v = k^ , pv"** = k2 = 
PiVjV. Whence the work done in expanding from Vi 
to v„ IS pdv = PjVjV — =z V r 1 = 

P V — P V 

-^^ — ^^ ^ , so that the work done by any gas in 

changing volume adiabatically is given by 



(55) W = Pl^L^-P^ = ^ilL fl . , 

r — 1 r — IL \v, 



An" Elementary Teeatise 167 

Also, since -^^ = E = ^r > ^-^-^ = -n^ > or, smce 

/ T \-i 

Pi^i^ = P2V y^2= ^1 f TpJ^ " ^ and, similarly, p^ = 

/ T \-^^^ c' 

Pi "TfT r ~ "^ J ill all of which y is the ratio — - = 1.408, 

and, if p is put in pounds per square foot and v in 
cubic feet, W will be the work in foot-pounds. 

(126) It must be noted that expressions (54) and 
(55) for work done in isothermal and in adiabatic ex- 
pansions are the total work done in changing volume, 
and that to find the available work it is necessary to 
subtract whatever work is done in overcoming the 
resistance of the back pressure. Moreover, in case 
the gas enters the working cylinder at a constant 
pressure, it does work by transmitted pressure, just 
as a liquid would, and this must be added to the work 
of expansion. 

Similarly, when work is done upon a gas, compress- 
ing it, the helping action of the back pressure must 
be taken account of, and so, too, must the work neces- 
sary to be done, after complete compression, in sweep- 
ing the compressed gas out of the cylinder. 

Thus suppose a volume v^ of gas at constant pressure 
Pi to enter a cylinder, and the supply to be then cut off 
and the gas allowed to expand adiabatically to volume 
V2 and pressure ps , after which the exhaust is opened, 
the volume v^ admitted on the other side of the piston, 
and the process reversed, the result being a reciprocat- 
ing motion which, by proper mechanical connections, 
performs useful work. Then at each stroke the 
gas does the work v^p^ in entering, and the work 

— ^^ — _ 1 in expanding, but during the whole 



168 Hydromechanics 

stroke the back pressure p2 , due to the gas being 
expelled from the other side of the cylinder, opposes 
the motion, and so the work VgPa is expended, leaving 

as the total available work v^p^ + _ -^ — V2P2 = 

Similarly, supposing the work done to consist in 
compressing a volume v^ of gas to V2 and then forcing 
it into a reservoir, then the work of compression is 

y p y p 

-^^ ~- , and the work of pushing the compressed 

gas out of the cylinder is V2P2 , but the back pressure 
Pi helps to do all this, and so reduces the expenditure 
of energy by v^p^ , making the total work done 

y^ (V2P2 — ViPi). 

If, however, the gas enters the working cylinder 
direct from a reservoir in which the pressure is not 
maintained, so that expansion takes place from the 
instant that admission of gas to the cylinder begins, 
then there is no work of entrance, but only the work 
of expansion given by (54) or (55) as the case may be, 
and in either case requiring to be reduced by the work 
done in overcoming the back pressure. 

Evidently it is unnecessary to consider each stroke 
of the piston separately, where the same cycle of 
operations is being repeated, and we can determine the 
whole available energy of a reservoir of compressed 
gas by applying our formula, taking v^ as the whole 
volume of the reservoir. 

(127) As an example, we will determine the avail- 
able energy per pound of air at 62° F. in a flask at 



Ax Elementary Treatise 169 

1500 pounds pressure per square inch. Here T-^ = 
522°; pi = 1500 X 144 = 216,000; and so v-, = 

53 3 T 

— ^ '= .1288 cu. ft. The air will give up all its 

Pi "^ ^ 

available energy when it is expanded to atmospheric 
pressure, so we have p2 = 14.7 X 144 = 2116, and 

P Y P V 

the work done in the expansion is ^^ y^— ^ which, 

since Vg = Y^l~j~y~ = 3.44, equals 50,345 foot-pounds. 

But of this work a portion, pg (v2 — v^) = 2116 X 
3.3112 = 7007 ft. lbs., is used in overcoming the 
atmospheric pressure during expansion, and so the 
total available energy is 50,345 — 7007 = 43,338 
ft. lbs. 

(128) As another example, we will take the work 
per stroke of a steam engine with piston area A, stroke 

a, and cut off at — the stroke, and we will assume the 
' n ' 

expansion to be isothermal, which is not far from the 

truth when wet steam is used. Then, if the entering 

pressure is m atmospheres, the work in entrance is 

aA 

— mpo , and the work of expanding to the end of the 

stroke is ~ loge n. But the work used in over- 

n ° 

coming the back pressure po is aAp^ . Therefore the 

useful work per stroke is aAp^ ( H log^ nj, 

Po being the atmospheric pressure. 

Problems XI. 

(1) The density of hydrogen is ^3^ that of air; what 
is the characteristic equation for hydrogen, and what 



170 Hydromechanics 

would be the pressure of 5 pounds of it contained in 
3 cubic feet and at 50° R? 

(2) What is the weight of the air in a room 20' by 
18' by 10', the temperature being 62° F. and the pres- 
sure 14.7 pounds per square inch? 

(3) The specific gravity of steam, compared with air, 
is .622. What is the characteristic equation for steam, 
and what volume of steam results from vaporizing 1 
cubic inch of water at 212° F. ? 

(4) A hydraulic ram 6" in diameter is supplied 
by a 2" pipe from an accumulator in which pressure 
is kept at 600 pounds per square inch. It takes 10^ 
of the accumulator pressure to overcome friction of the 
mechanical parts and the total hydraulic resistance 
is 10 times the velocity head in the supply pipe when 
a force of 12,000 pounds is being exerted by the ram. 
What is the speed of steady motion? If the speed is 
doubled, what force is being exerted by the ram? 

(5) If in the preceding example the partial closing 
of the supply valve doubles the hydraulic resistance, 
what then is the speed of steady motion? 

(6) A weight of 10 tons is moving 5 f. s. and it is 
desired to reduce its speed to 1 f. s. in a travel of 1' 
by means of a hydraulic brake having a piston of 16 
square inches effective area. What must the escape 
area from one side to the other of the piston be? 

(7) If a hydraulic brake is to oppose constant resist- 
ance to motion, so as to uniformly retard a weight W 
which starts with velocity V and to bring it to rest 
in the distance 1, deduce a formula for the escape area 
a, the piston area being A. 

(8) Air at 62° F. and 15 pounds pressure per square 
inch is suddenly compressed to one-fifth its volume; 



A^ Elementary Teeatise 171 

what do its temperature and pressure become ? What 
will the pressure be when it is cooled again to 62° F. ? 
What will its temperature be if, after cooling, it is 
allowed to expand adiabatically to its original pressure ? 

(9) Compare the work done in compressing each 
pound of air as in example (8) with the work it will 
do, after cooling, in again expanding to atmospheric 
pressure. 

(10) Compare the work of obtaining a pound of air 
at 4 atmospheres pressure and atmospheric tempera- 
ture (62° F.) by compression adiabatically and iso- 
thermally. 

(11) The cylinder of a compressed air engine is of 1 
square foot cross section and the stroke is 18''. If 
the initial pressure is 3 atmospheres and the tempera- 
ture 62° F., what should be the point of cut-off so 
that the pressure may be atmospheric at the end of 
the stroke? What then is the available work done 
per stroke, and also per pound of air, and what is the 
temperature of the exhaust air? 

(12) Compare the work of obtaining a pound of air 
at 100 atmospheres pressure and at atmospheric pres- 
sure by one compression, and by three successive equal 
compressions with time for cooling the air between 
them. 

(13) The air flask of a 5-meter Whitehead torpedo 
contains 11.77 cubic feet of air at 1500 pounds per 
square inch. What is its available energy? 

(14) If the air of example (13) is used at a constant 
pressure of 300 pounds per square inch by means of a 
reducing valve, what is the available energy? 

(15) If in example (14) the expansion in the work- 
ing cylinders of the engine is incomplete, the cut-off 



173 Hydromechanics 

being at one-third the stroke, what is the available 
energy ? 

(16) A pound of gunpowder in burning forms -^ 
pounds of gas and gives up 1400 B. T. units. Suppos- 
ing the solid and liquid residue to give all their heat 
to the gun walls, so that the gas expands adiabatically, 
how many foot-tons of energy will a pound of powder 
supply in expanding three times? 

(17) If the combustion of a pound of smokeless 
powder converts it all into gas and develops 2000 B. T. 
units, how many foot-tons of work would be done per 
pound of powder in expanding four times, and with 
that expansion in a 6" gun, what velocity would 25 
pounds of such powder give to a 100-pound projectile ? 

(18) Prove that in supplying a reservoir with com- 
pressed air by machinery which does the compression 
in n stages, with complete cooling after each stage, it 
is most economical to do the same amount of work 
at each compression, and, if k is the ratio of initial 
to final volume, to reduce the volume in the ratio ^~k 
at each compression. 



An Elementary Treatise 173 



ANSWEES TO PEOBLEMS. 

Problems I.— (1) 288 pounds. (3) 3.86 pounds. 
(3) 491.07 pounds. (4) 736.6 pounds. (5) + 4 pounds. 
(6) 900 pounds. (7) 50 pounds. (8) 2592 pounds. 
(9). 286.4 pounds per sq. in. (10) 10,056 tons. (11) 
1678 pounds per sq. in. (12) 110.2 cu. in., 16.53 
pounds per sq. in. (13) 2369 cu. yds. (14) 40.12 
pounds per sq. in.; y^ = 0.497Vo . (15) 300 and 
1018.5 pounds per sq. in. (16) Spherical. (17) 91.3 
pounds, 1201 cu. ft. (18) About 400,000 ft. lbs. 

Problems II.— (1) 32.36 pounds; 4660 pounds. (2) 
33.'87; 33.'08. (3) 2666.7 pounds per square inch. 
(5) 360 pounds. (6) 5625 pounds. (7) 0.542 pounds 
per sq. in. (8) 0.276 pounds per sq. in. (10) 2 + \/Y. 
(11) 106f ft. (12) Vi = 2.201Vo. (13) y^ = 

y 

:^ — ; — rl^A-i — j> whcTC X is depth in inches of water 
1 + .00241 X ^ ^ 



surface in cone. (14) y, = [yI^T^^^ ^ • ^^^) 
6934'. (17) 23876'. (18) 27613'. (19) 32.'28. (20) 
(a) It TYiLl still just float (neglecting a very small in- 
crease of weight of euYclope due to decreased density 
of air it displaces), (b) It would sink. (21) (h — 
414.72)A cu. in. 

Problems IIP— {1) 2:1. (2) 62,071 pounds. (3) 
1 : 2. (4) At its mid height. (5) So as to cut off f 
the opposite side. (6) 6'.'13 in the water. (7) Upper 
faces equal and each half that on each lower face. 
(8) 50,000 pounds. (9) 19'.76. (11) 1:15.6. (12) 7'. 



174 Hydromechanics 



(14) The lower corner supports ^ and the others each J. 

4h 

(15) y . (16) 346| pounds. (17) 13/44 and 10/38. 

(18) 9.'43 and 7.'26. (19) 1" below its center. (20) fa 

7ab 
below surface. (21) - — , „ , „ below surface. (22) 
^ ^ 6 s/'^ + b^ 

a^ 
~jr- , where h is depth of center. (24) h = fh^ , 

where h^ is barometric height. (25) Half the height 
of the cone. 

ProUems IV.— (1) 0.4903 pounds. (2) 3.144 cu. ft. 
(3) 5.926 cu. ft. (4) 417.5 and 370 pounds. (5) 11.44. 
(6) 21.505 and 0.914. (7) 0.771. (8) 147 pounds. 
(9) 19.5. (10) 18.41. (11) 1.0952, 0.9747; 1.0000. 
(12) 0.87 copper, 0.13 tin. (13) 20. (14) 0.2516 cu. ft. 
(15) a(l — l-v^S): (16) .03118 cu. ft. or 53.87 cu. in. 
(17) 3.5 grains. (18) 2.5955 silver to 1 copper; or 
11.55 to 4.45. (19) 11:9 weight, 2:3 vol. (20) wrh^ 

acting — below top. (21) wr^h (2 4- -^ ) acting through 
wrh 



center of axis. (22) — - — \/9h^ + Tr^r^ making 

tan~^ — with axis. (24) ^--— wrr^ through center. 

wrrr^ 



(25) -^ v'13 — 12 cos and -^ ^13 + 12 cos 0. 

(26) 443 f. s.; 13.755 pounds per sq. in.; 1830'; 3384'; 

inside same as before; outside, 12.95 pounds per sq. in. 

/«.v^ W7rr^ / , lor sin 2'x \ ^ 5w7rr^ cos^ a ,^„^ 
(^V ^- [^ g j and . (28) 

3^10 • ^^^^ T' 



An Elementary Treatise 175 

Problems V.—{1) 2.13 f. s.; 5496'; 1680'; 25 pounds. 
(2) 135.86 pounds; 2298'; 9180'. (3) Draft is 

h (l- '^1--^]. (4)-^. (5) 2388 pounds. (8) 
h ll-Jl —]. (9) Just immersed. (10) 4267 

tons. (11) 24' — 2'.'46; 293 tons. (15) h. = ^^ . 
(19) p > sec* a, when a is semi-vertical angle. (21) 

4 
ProUems VI.— (1) ^ . (2) a>^ = ^ . (3) w' = 

4g ( -^ - V J 22 a>2 

\ ^ L . (4) a - -J vertically. (5) — (OA^ _ 

OB^). (6) -^ w; -1'^ (-4^ + hj w; and ;rrh [~^ + 

gh 
li)w. (7) It will be emptied by rotation w^ = - — 

where a is distance from axis of cone to axis of 
rotation. (8) 23'.04; 5'.06; 28'.10. (9) 12.8 f. s. and 
7.11 f. s.; 2'.56 and 0'.79; 3'.44 and 4'.21; 2.60 and 2.17 
pounds per sq. in.; 1.49 and 1.83 pounds per sq. in. 
(10) 0.98 pounds per sq. in. below atmospheric pres- 
sure. (11) 66.51 f. s.; 4.24 pounds per sq. in. (12) 5.66 
f.s.; 17.89 f.s.; 80 f. s. (13) 46.99 f. s.; 49.96 f. s.; 
92.61 f.s. (14) 121.56 f.s.; 122.75 f.s.; 145.42 f.s. 
(15) 56.57 f.s. (16) 100.5 f.s. (17) 121.96 f.s. 

(18) 42.64 f. s. (19) 14.73 f. s. (25) to' = -^ . 

ProUems VII.— (1) 363.7 cu. ft. (2) 1'.232. (3) 366 

^, ,,^ 4cA 8c A 37rcA 

cu.ft. (4)--,/2ga. (5) -^yx/2ga. (6) -.^g- V^ga . 



IVIAR 15 1902 

1 COPY DEL. TO CAT. I rv. 
r4AR. 15 1902 



^AR. \S 1902 



